Tüm alıştırma soruları

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Soru 161Soru →

In a direct current (DC) avionics test circuit, a 48 V48\text{ V} power supply is connected in series with a 30 Ω30\text{ }\Omega current-limiting resistor and a 50 Ω50\text{ }\Omega load resistor. What is the electrical power dissipated by the 50 Ω50\text{ }\Omega load resistor, in watts?

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Cevap: 18

Cevap

The power dissipated by the 50 Ω50\text{ }\Omega load resistor is 18 W18\text{ W}.
In a series circuit, the total resistance is the sum of all individual resistances (30 Ω+50 Ω=80 Ω30\text{ }\Omega + 50\text{ }\Omega = 80\text{ }\Omega). With a 48 V48\text{ V} supply, the circuit current is I=48 V80 Ω=0.6 AI = \frac{48\text{ V}}{80\text{ }\Omega} = 0.6\text{ A}. The electrical power dissipated by the 50 Ω50\text{ }\Omega resistor is determined by P=I2R=(0.6 A)2×50 Ω=18 WP = I^2 R = (0.6\text{ A})^2 \times 50\text{ }\Omega = 18\text{ W}.

Adım Adım Çözüm

1
Find the equivalent total resistance of the series circuit
Rtotal=30 Ω+50 Ω=80 ΩR_{\text{total}} = 30\text{ }\Omega + 50\text{ }\Omega = 80\text{ }\Omega
Resistances in a series circuit are directly additive to determine total opposition to current flow.
2
Calculate the total circuit current using Ohm's Law
I=VtotalRtotal=48 V80 Ω=0.6 AI = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{48\text{ V}}{80\text{ }\Omega} = 0.6\text{ A}
Current is uniform throughout all components connected in a series loop.
3
Calculate the power dissipation in the load resistor using Watt's Law
P=I2×R=(0.6 A)2×50 Ω=0.36 A2×50 Ω=18 WP = I^2 \times R = (0.6\text{ A})^2 \times 50\text{ }\Omega = 0.36\text{ A}^2 \times 50\text{ }\Omega = 18\text{ W}
Power dissipated by a resistor is proportional to the square of the current flowing through it multiplied by its resistance.

Anahtar Kavram

Series circuit current and resistor power dissipation using combined Ohm's Law and Watt's Law (P=I2RP = I^2 R or P=V2RP = \frac{V^2}{R})

Alternatif Yöntem

Find the voltage drop across the load resistor using the voltage divider rule: Vload=48 V×(50 Ω30 Ω+50 Ω)=48×58=30 VV_{\text{load}} = 48\text{ V} \times \left(\frac{50\text{ }\Omega}{30\text{ }\Omega + 50\text{ }\Omega}\right) = 48 \times \frac{5}{8} = 30\text{ V}. Then compute power using P=Vload2Rload=30250=90050=18 WP = \frac{V_{\text{load}}^2}{R_{\text{load}}} = \frac{30^2}{50} = \frac{900}{50} = 18\text{ W}.
Tahmini Süre:1m 15s
Soru 162Soru →

A railway signal monitoring circuit is powered by a 72 V72\text{ V} DC power supply. The circuit consists of three components connected in series: an 85 Ω85\ \Omega relay coil, a 35 Ω35\ \Omega surge-limiting resistor, and an unknown protective resistor RxR_x. If the entire circuit draws a steady current of 0.4 A0.4\text{ A}, what is the resistance of the protective resistor RxR_x in ohms (Ω\Omega)?

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Cevap: 60

Cevap

The resistance of the protective resistor RxR_x is 60 Ω60\ \Omega.
Applying Ohm's Law (RT=VTIR_T = \frac{V_T}{I}) gives a total series resistance of 180 Ω180\ \Omega. Since resistances in series add directly (RT=85 Ω+35 Ω+Rx=120 Ω+RxR_T = 85\ \Omega + 35\ \Omega + R_x = 120\ \Omega + R_x), solving for the unknown yields Rx=180 Ω−120 Ω=60 ΩR_x = 180\ \Omega - 120\ \Omega = 60\ \Omega.

Adım Adım Çözüm

1
Calculate the total circuit resistance (RTR_T) using Ohm's Law.
RT=VsourceItotal=72 V0.4 A=180 ΩR_T = \frac{V_{\text{source}}}{I_{\text{total}}} = \frac{72\text{ V}}{0.4\text{ A}} = 180\ \Omega
In a series circuit, total resistance equals total source voltage divided by total circuit current.
2
Determine the combined resistance of the known series components.
Rknown=Rrelay+Rsurge=85 Ω+35 Ω=120 ΩR_{\text{known}} = R_{\text{relay}} + R_{\text{surge}} = 85\ \Omega + 35\ \Omega = 120\ \Omega
Resistances in a series circuit are directly additive.
3
Subtract the known series resistance from the total circuit resistance to solve for RxR_x.
Rx=RT−Rknown=180 Ω−120 Ω=60 ΩR_x = R_T - R_{\text{known}} = 180\ \Omega - 120\ \Omega = 60\ \Omega
The sum of all individual series resistors must equal the total equivalent resistance (RT=R1+R2+RxR_T = R_1 + R_2 + R_x).

Anahtar Kavram

In a direct-current (DC) series circuit, current is identical through every component (IT=I1=I2=I3I_T = I_1 = I_2 = I_3), and the total equivalent resistance is the arithmetic sum of all individual resistances (RT=R1+R2+⋯+RnR_T = R_1 + R_2 + \dots + R_n).

Alternatif Yöntem

Find the voltage drops across the known components first: Vrelay=0.4 A×85 Ω=34 VV_{\text{relay}} = 0.4\text{ A} \times 85\ \Omega = 34\text{ V} and Vsurge=0.4 A×35 Ω=14 VV_{\text{surge}} = 0.4\text{ A} \times 35\ \Omega = 14\text{ V}. Subtract these from the total source voltage using Kirchhoff's Voltage Law: VRx=72 V−(34 V+14 V)=24 VV_{Rx} = 72\text{ V} - (34\text{ V} + 14\text{ V}) = 24\text{ V}. Then find Rx=VRxI=24 V0.4 A=60 ΩR_x = \frac{V_{Rx}}{I} = \frac{24\text{ V}}{0.4\text{ A}} = 60\ \Omega.
Tahmini Süre:1m 0s
Soru 163Soru →

A mobile radar subsystem is powered by a 24 V24\text{ V} direct-current source. Its internal circuitry contains a parallel configuration with three distinct branches.

The total equivalent resistance of this parallel section is exactly 3 Ω3\ \Omega.
A technician measures the current through the first branch to be 3 A3\text{ A}.
The second branch has a known resistance of 24 Ω24\ \Omega.

What is the resistance of the third branch in ohms?

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Cevap: 6

Cevap

The resistance of the third branch is 6 Ω6\ \Omega.
The correct value is derived by finding the total circuit current (8 A8\text{ A}) and subtracting the known branch currents (3 A3\text{ A} and 1 A1\text{ A}) to find the current in the third branch (4 A4\text{ A}). Dividing the common 24 V24\text{ V} source by this 4 A4\text{ A} current yields 6 Ω6\ \Omega.

Adım Adım Çözüm

1
Calculate the total circuit current using the total voltage and equivalent resistance.
Itotal=24 V/3 Ω=8 AI_{total} = 24\text{ V} / 3\ \Omega = 8\text{ A}
Ohm's law applies to the entire circuit, linking total voltage, total current, and equivalent resistance.
2
Determine the current flowing through the second branch.
I2=24 V/24 Ω=1 AI_2 = 24\text{ V} / 24\ \Omega = 1\text{ A}
In a parallel circuit, the full source voltage drops across every branch. Ohm's law (I=V/RI = V/R) gives the branch current.
3
Calculate the current flowing through the third branch.
I3=8 A−(3 A+1 A)=4 AI_3 = 8\text{ A} - (3\text{ A} + 1\text{ A}) = 4\text{ A}
Kirchhoff's Current Law states that the total current is the sum of the individual branch currents (Itotal=I1+I2+I3I_{total} = I_1 + I_2 + I_3).
4
Calculate the resistance of the third branch.
R3=24 V/4 A=6 ΩR_3 = 24\text{ V} / 4\text{ A} = 6\ \Omega
Apply Ohm's law (R=V/IR = V/I) specifically to the third branch using the source voltage and the calculated branch current.

Anahtar Kavram

Applying Ohm's law to both individual branches and the entire circuit, combined with parallel circuit rules for voltage and current.

Alternatif Yöntem

Use the parallel resistance reciprocal formula: 1/Req=1/R1+1/R2+1/R31/R_{eq} = 1/R_1 + 1/R_2 + 1/R_3. First, find R1=V/I1=24 V/3 A=8 ΩR_1 = V/I_1 = 24\text{ V} / 3\text{ A} = 8\ \Omega. Then, substitute the known values: 1/3=1/8+1/24+1/R31/3 = 1/8 + 1/24 + 1/R_3. Finding a common denominator of 24 gives 8/24=3/24+1/24+1/R38/24 = 3/24 + 1/24 + 1/R_3, which simplifies to 8/24=4/24+1/R38/24 = 4/24 + 1/R_3. Therefore, 1/R3=4/24=1/61/R_3 = 4/24 = 1/6, meaning R3=6 ΩR_3 = 6\ \Omega.
Tahmini Süre:1m 30s
Soru 164Soru →

An avionics technician is servicing a step-down transformer in an aircraft subsystem. The transformer receives an input voltage of 115 V AC115\text{ V AC} across its primary coil and delivers an output voltage of 23 V AC23\text{ V AC} across its secondary coil. If the primary winding has 750 turns750\text{ turns}, what is the number of turns on the secondary winding?

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Cevap: 150

Cevap

150 turns
The relationship between primary and secondary voltage and turns in a transformer is VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s}. Given a primary voltage of 115 V AC115\text{ V AC} and a secondary voltage of 23 V AC23\text{ V AC}, the transformer has a step-down ratio of 5:15:1 (115/23=5115 / 23 = 5). Dividing the primary turns (750750) by 55 gives 150 turns150\text{ turns} for the secondary winding.

Adım Adım Çözüm

1
State the fundamental transformer turns ratio formula relating voltage and winding turns.
VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s}
The voltage induced across transformer windings is directly proportional to the number of turns on each winding.
2
Isolate the secondary turns variable (NsN_s).
Ns=Np×VsVpN_s = N_p \times \frac{V_s}{V_p}
Algebraic rearrangement allows direct calculation of the secondary coil turns.
3
Substitute the given values (Vp=115 V ACV_p = 115\text{ V AC}, Vs=23 V ACV_s = 23\text{ V AC}, Np=750 turnsN_p = 750\text{ turns}) and simplify.
Ns=750×23115=750×0.2=150 turnsN_s = 750 \times \frac{23}{115} = 750 \times 0.2 = 150\text{ turns}
The voltage reduction factor is 5:15:1, meaning the secondary winding must have one-fifth the number of turns of the primary winding.

Anahtar Kavram

Transformer turns and voltage ratio relationship
Tahmini Süre:1m 0s
Soru 165Soru →

A field communication station is powered by a 72 V72\text{ V} direct-current supply connected across three parallel load branches. The resistance of the first branch is 18 Ω18\ \Omega, the resistance of the second branch is 36 Ω36\ \Omega, and the resistance of the third branch is 72 Ω72\ \Omega. What is the total current, in amperes (A\text{A}), supplied to the circuit?

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Cevap: 7

Cevap

7 A
Each of the three branches is connected directly across the 72 V72\text{ V} source. Using Ohm's Law (I=V/RI = V/R), the branch currents are 4 A4\text{ A}, 2 A2\text{ A}, and 1 A1\text{ A}. Adding these currents together yields the total current of 7 A7\text{ A}.

Adım Adım Çözüm

1
Calculate the current in the first parallel branch using I1=VR1I_1 = \frac{V}{R_1}.
I1=72 V18 Ω=4 AI_1 = \frac{72\text{ V}}{18\ \Omega} = 4\text{ A}
In a parallel DC circuit, the full supply voltage of 72 V72\text{ V} appears across every parallel branch.
2
Calculate the current in the second parallel branch using I2=VR2I_2 = \frac{V}{R_2}.
I2=72 V36 Ω=2 AI_2 = \frac{72\text{ V}}{36\ \Omega} = 2\text{ A}
Applying Ohm's Law to the second branch with its specific resistance.
3
Calculate the current in the third parallel branch using I3=VR3I_3 = \frac{V}{R_3}.
I3=72 V72 Ω=1 AI_3 = \frac{72\text{ V}}{72\ \Omega} = 1\text{ A}
Applying Ohm's Law to the third branch with its specific resistance.
4
Add all three individual branch currents together to determine the total current.
Itotal=4 A+2 A+1 A=7 AI_{\text{total}} = 4\text{ A} + 2\text{ A} + 1\text{ A} = 7\text{ A}
By Kirchhoff's Current Law, the total current leaving the power supply equals the sum of the currents through all parallel branches.

Anahtar Kavram

In a parallel circuit, the voltage across each parallel branch is identical to the source voltage, and the total circuit current is the sum of the individual branch currents (Itotal=I1+I2+⋯+InI_{\text{total}} = I_1 + I_2 + \dots + I_n).

Alternatif Yöntem

Find the total equivalent resistance (ReqR_{\text{eq}}) first: 1Req=118+136+172=472+272+172=772 Ω−1\frac{1}{R_{\text{eq}}} = \frac{1}{18} + \frac{1}{36} + \frac{1}{72} = \frac{4}{72} + \frac{2}{72} + \frac{1}{72} = \frac{7}{72}\ \Omega^{-1}, so Req=727 ΩR_{\text{eq}} = \frac{72}{7}\ \Omega. Then find total current using Itotal=VReq=7272/7=7 AI_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{72}{72/7} = 7\text{ A}.
Tahmini Süre:1m 0s
Soru 166Soru →

A radio technician is testing the output of an RF signal generator using an oscilloscope. The horizontal timebase control is set to 250 μs/div250\text{ }\mu\text{s/div}. On the screen graticule, a continuous sinusoidal waveform is observed where exactly 2.5 complete cycles2.5\text{ complete cycles} span across 8 horizontal divisions8\text{ horizontal divisions}. What is the frequency of this signal, in hertz (Hz\text{Hz})?

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Cevap: 1250

Cevap

1250 Hz
The total duration of the 8 horizontal divisions8\text{ horizontal divisions} at 250 μs/div250\text{ }\mu\text{s/div} is 2000 μs2000\text{ }\mu\text{s} (2 ms2\text{ ms}). Dividing by the 2.5 complete cycles2.5\text{ complete cycles} gives a period of T=800 μs=0.0008 sT = 800\text{ }\mu\text{s} = 0.0008\text{ s} per cycle. Taking the reciprocal gives the frequency: f=1/0.0008 s=1250 Hzf = 1 / 0.0008\text{ s} = 1250\text{ Hz}.

Adım Adım Çözüm

1
Calculate the total elapsed time across the observed graticule divisions.
Total time =8 div×250 μs/div=2000 μs= 8\text{ div} \times 250\text{ }\mu\text{s/div} = 2000\text{ }\mu\text{s}
The horizontal axis represents time, and the timebase setting indicates the duration per division.
2
Calculate the period (TT) of a single cycle.
T=2000 μs2.5 cycles=800 μs=800×10−6 s=0.0008 sT = \frac{2000\text{ }\mu\text{s}}{2.5\text{ cycles}} = 800\text{ }\mu\text{s} = 800 \times 10^{-6}\text{ s} = 0.0008\text{ s}
The period is the time required to complete one single cycle.
3
Calculate the frequency (ff) from the period.
f=1T=10.0008 s=1250 Hzf = \frac{1}{T} = \frac{1}{0.0008\text{ s}} = 1250\text{ Hz}
Frequency in hertz is the reciprocal of the period in seconds (f=1/Tf = 1/T).

Anahtar Kavram

Calculating waveform period and frequency from oscilloscope horizontal timebase settings and graticule divisions.
Tahmini Süre:1m 30s
Soru 167Soru →

An oscilloscope is used to measure an alternating signal. The horizontal sweep rate (Time/Div) is set to 8 μs/div8\text{ }\mu\text{s/div}, and one complete cycle of the waveform spans exactly 5 horizontal divisions5\text{ horizontal divisions} across the graticule. What is the frequency of this signal in kilohertz (kHz\text{kHz})?

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Cevap: 25

Cevap

The frequency of the signal is 25 kHz25\text{ kHz}.
One full waveform cycle occupies 55 horizontal divisions at 8 μs/division8\text{ }\mu\text{s/division}, giving a period of T=40 μsT = 40\text{ }\mu\text{s} (40×10−6 s40 \times 10^{-6}\text{ s}). The frequency is the reciprocal of the period (f=1T=140×10−6 s=25000 Hzf = \frac{1}{T} = \frac{1}{40 \times 10^{-6}\text{ s}} = 25{}000\text{ Hz}), which equals 25 kHz25\text{ kHz}.

Adım Adım Çözüm

1
Multiply the number of horizontal graticule divisions per cycle by the time base setting to find the period (TT).
T=5 divisions×8 μs/div=40 μs=40×10−6 sT = 5\text{ divisions} \times 8\text{ }\mu\text{s/div} = 40\text{ }\mu\text{s} = 40 \times 10^{-6}\text{ s}
Horizontal divisions represent the time dimension on an oscilloscope display.
2
Calculate the frequency (ff) in hertz using the reciprocal relationship f=1Tf = \frac{1}{T}.
f=140×10−6 s=25000 Hzf = \frac{1}{40 \times 10^{-6}\text{ s}} = 25{}000\text{ Hz}
Frequency is defined as the number of cycles per second.
3
Convert the resulting frequency from hertz (Hz\text{Hz}) to kilohertz (kHz\text{kHz}).
25000 Hz1000=25 kHz\frac{25{}000\text{ Hz}}{1{}000} = 25\text{ kHz}
The question specifies the final answer unit as kilohertz.

Anahtar Kavram

Calculating waveform period and frequency from oscilloscope horizontal Time/Div settings and graticule division counts
Tahmini Süre:1m 0s
Soru 168Soru →

An industrial direct-current (DC) indicator lamp dissipates 1.8 W1.8\text{ W} of power when a current of 150 mA150\text{ mA} flows through it. What is the voltage drop across the lamp, in volts?

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Cevap: 12

Cevap

The voltage drop across the indicator lamp is 12 V12\text{ V}.
According to Watt's Law, electrical power is the product of voltage and current (P=V×IP = V \times I). Rearranging the formula to solve for voltage gives V=PIV = \frac{P}{I}. Converting 150 mA150\text{ mA} to 0.15 A0.15\text{ A} and evaluating 1.8 W0.15 A\frac{1.8\text{ W}}{0.15\text{ A}} yields 12 V12\text{ V}.

Adım Adım Çözüm

1
Convert the current from milliamperes to amperes.
I=150 mA=0.15 AI = 150\text{ mA} = 0.15\text{ A}
Standard electrical formulas require current in base units of amperes (A).
2
Rearrange Watt's Law to solve for voltage.
V=PIV = \frac{P}{I}
Watt's Law defines power as P=V×IP = V \times I, so isolating voltage yields V=PIV = \frac{P}{I}.
3
Substitute the known values into the equation to find the voltage.
V=1.8 W0.15 A=12 VV = \frac{1.8\text{ W}}{0.15\text{ A}} = 12\text{ V}
Dividing the power dissipation (1.8 W1.8\text{ W}) by the current (0.15 A0.15\text{ A}) gives the operating voltage.

Anahtar Kavram

Watt's Law describes the mathematical relationship between electrical power, voltage, and current: P=V×IP = V \times I. When power and current are known, voltage is calculated using V=PIV = \frac{P}{I}, provided current is expressed in amperes.
Tahmini Süre:1m 0s
Soru 169Soru →

An electronic circuit uses a standard four-band resistor with the color bands Yellow, Violet, Brown, and Gold reading from left to right. What is the nominal resistance value of this resistor in ohms (Ω\Omega)?

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Cevap: 470

Cevap

The nominal resistance value is 470 Ω470\ \Omega.
In the standard electronic color code system, Yellow represents 4, Violet represents 7, Brown represents a multiplier of 10 (10110^1), and Gold indicates a tolerance of ±5%\pm 5\%. Combining the digits gives 47, and multiplying by 10 produces 470 Ω470\ \Omega.

Adım Adım Çözüm

1
Decode the first two color bands into the first two digits
Yellow corresponds to 4 and Violet corresponds to 7, giving the base number 47
The first two bands on a four-band resistor represent the first and second significant figures of the resistance value
2
Decode the third band into the decimal multiplier
Brown corresponds to a multiplier of 10110^1 (1010)
The third band indicates the power of ten by which the base number must be multiplied
3
Calculate the total nominal resistance
47×10=470 Ω47 \times 10 = 470\ \Omega
Multiplying the base digits by the multiplier gives the rated nominal resistance in ohms

Anahtar Kavram

Four-band resistor color code decoding
Tahmini Süre:30s
Soru 170Soru →

A maintenance technician is tracking the number of work orders completed each day. Over a 55-day work week, the technician wants to achieve a mean of 1414 completed work orders per day. If the technician completed 1212, 1616, 1111, and 1313 work orders on the first four days, how many work orders must be completed on the fifth day to exactly meet the target mean?

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Cevap: 18

Cevap

The technician must complete 18 work orders on the fifth day.
To find a missing value when the mean is known, first determine the total sum required by multiplying the target mean (14) by the total number of data points (5), which equals 70. Next, add the known values from the first four days (12 + 16 + 11 + 13 = 52). Finally, subtract the sum of the known values from the total required sum (70 - 52 = 18). Therefore, 18 work orders must be completed on the fifth day.

Adım Adım Çözüm

1
Determine the total sum required for all 5 days.
14×5=7014 \times 5 = 70
The mean is the total sum divided by the number of values. To find the required total sum, multiply the target mean by the total number of days.
2
Calculate the sum of the work orders completed so far.
12 + 16 + 11 + 13 = 52
This gives the total number of work orders already completed in the first four days.
3
Find the difference between the required total and the completed total.
70 - 52 = 18
Subtracting the sum of the first four days from the required 5-day total yields the exact value needed for the fifth day.

Anahtar Kavram

Calculating a missing value from a given mean requires multiplying the mean by the total number of items to find the required sum, then subtracting the known values.
Soru 171Soru →

An electronics technician is fabricating a heating coil using nichrome resistance wire. The wire has a cross-sectional area of 0.2 mm20.2\text{ mm}^2, a total length of 6 meters6\text{ meters}, and a material resistivity of 1.10 Ω⋅mm2/m1.10\ \Omega\cdot\text{mm}^2/\text{m}. What is the total electrical resistance of the heating coil in ohms (Ω\Omega) across its entire length?

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Cevap: 33

Cevap

The total resistance of the nichrome wire heating coil is 33 Ω33\ \Omega.
The total resistance of any conductor is governed by R=ρLAR = \rho \frac{L}{A}. Substituting the length of 6 m6\text{ m}, cross-sectional area of 0.2 mm20.2\text{ mm}^2, and resistivity of 1.10 Ω⋅mm2/m1.10\ \Omega\cdot\text{mm}^2/\text{m} yields R=1.10×60.2=33 ΩR = \frac{1.10 \times 6}{0.2} = 33\ \Omega.

Adım Adım Çözüm

1
State the governing physical formula relating electrical resistance to conductor dimensions and material resistivity.
R=ρLAR = \rho \frac{L}{A}, where RR is resistance in ohms, ρ\rho is resistivity, LL is length, and AA is cross-sectional area.
Electrical resistance is directly proportional to wire length and resistivity, and inversely proportional to cross-sectional area.
2
Substitute the given numerical values into the formula with compatible units.
R=1.10 Ω⋅mm2/m×6 m0.2 mm2R = \frac{1.10\ \Omega\cdot\text{mm}^2/\text{m} \times 6\text{ m}}{0.2\text{ mm}^2}
The unit of resistivity (Ω⋅mm2/m\Omega\cdot\text{mm}^2/\text{m}) aligns directly with meters for length and square millimeters for area, canceling out spatial units.
3
Calculate the numerator product and divide by the cross-sectional area.
R=6.600.2=33 ΩR = \frac{6.60}{0.2} = 33\ \Omega
Dividing 6.606.60 by 0.20.2 gives the final total resistance in ohms.

Anahtar Kavram

Calculation of electrical resistance from material resistivity, conductor length, and cross-sectional area (R=ρL/AR = \rho L / A)
Tahmini Süre:1m 0s
Soru 172Soru →

A marine auxiliary direct-current power circuit supplies a constant 48 V48\text{ V} across three parallel branches. The first branch has a resistance of 16 Ω16\ \Omega, the second branch has a resistance of 24 Ω24\ \Omega, and the third branch has a resistance of 48 Ω48\ \Omega. What is the total current, in amperes (A\text{A}), drawn from the power supply?

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Cevap: 6

Cevap

The total current drawn from the power supply is 6 A6\text{ A}.
Each parallel branch receives the full source voltage of 48 V48\text{ V}. Dividing the voltage by each branch resistance yields branch currents of 3 A3\text{ A}, 2 A2\text{ A}, and 1 A1\text{ A}. Summing these branch currents gives a total circuit current of 6 A6\text{ A}.

Adım Adım Çözüm

1
Calculate the current passing through the first branch
I1=48 V16 Ω=3 AI_1 = \frac{48\text{ V}}{16\ \Omega} = 3\text{ A}
The full source voltage of 48 V48\text{ V} appears across every parallel branch.
2
Calculate the current passing through the second branch
I2=48 V24 Ω=2 AI_2 = \frac{48\text{ V}}{24\ \Omega} = 2\text{ A}
Applying Ohm's Law (I=VRI = \frac{V}{R}) with the same supply voltage across the 24 Ω24\ \Omega branch.
3
Calculate the current passing through the third branch
I3=48 V48 Ω=1 AI_3 = \frac{48\text{ V}}{48\ \Omega} = 1\text{ A}
Applying Ohm's Law (I=VRI = \frac{V}{R}) across the 48 Ω48\ \Omega branch.
4
Sum all individual branch currents to determine total current
Itotal=3 A+2 A+1 A=6 AI_{\text{total}} = 3\text{ A} + 2\text{ A} + 1\text{ A} = 6\text{ A}
In a parallel circuit, the total current equals the sum of all branch currents (Itotal=I1+I2+I3I_{\text{total}} = I_1 + I_2 + I_3).

Anahtar Kavram

In a parallel DC circuit, the voltage across each branch is identical to the source voltage, and the total current supplied is the sum of the individual branch currents.

Alternatif Yöntem

Find the equivalent resistance (ReqR_{\text{eq}}) first using the reciprocal formula: 1Req=116+124+148=348+248+148=648=18  ⟹  Req=8 Ω\frac{1}{R_{\text{eq}}} = \frac{1}{16} + \frac{1}{24} + \frac{1}{48} = \frac{3}{48} + \frac{2}{48} + \frac{1}{48} = \frac{6}{48} = \frac{1}{8}\implies R_{\text{eq}} = 8\ \Omega. Then, calculate total current via Ohm's Law: Itotal=VReq=48 V8 Ω=6 AI_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{48\text{ V}}{8\ \Omega} = 6\text{ A}.
Tahmini Süre:1m 0s
Soru 173Soru →

A solid copper conductor with a length of 150 meters150\text{ meters} and a uniform cross-sectional area of 3.0 mm23.0\text{ mm}^2 has a measured resistance of 0.85 Ω0.85\ \Omega. An electrical technician replaces this conductor with another copper conductor of the same composition having a length of 90 meters90\text{ meters} and a cross-sectional area of 1.5 mm21.5\text{ mm}^2. What is the electrical resistance of the replacement conductor, in ohms (Ω\Omega)?

Cevabı ve açıklamayı göster

Cevap: 1.02

Cevap

1.02 ohms (Ω\Omega)
Resistance depends directly on conductor length and inversely on cross-sectional area. The length decreases by a factor of 90150=0.60\frac{90}{150} = 0.60, while the cross-sectional area decreases by a factor of 1.53.0=0.50\frac{1.5}{3.0} = 0.50, which doubles the resistance contribution (factor of 2.02.0). Multiplying the original resistance by both factors gives 0.85 Ω×0.60×2.0=1.02 Ω0.85\ \Omega \times 0.60 \times 2.0 = 1.02\ \Omega.

Adım Adım Çözüm

1
State the governing formula for electrical resistance of a uniform conductor.
R=ρLAR = \rho \frac{L}{A}, where ρ\rho is material resistivity, LL is conductor length, and AA is cross-sectional area.
Since both conductors are made of identical copper, resistivity ρ\rho remains constant, making resistance directly proportional to length and inversely proportional to cross-sectional area.
2
Determine the scaling factor resulting from the change in length.
Length factor =L2L1=90 m150 m=0.60= \frac{L_2}{L_1} = \frac{90\text{ m}}{150\text{ m}} = 0.60.
Decreasing conductor length proportionally reduces resistance to 60% of its initial value.
3
Determine the scaling factor resulting from the change in cross-sectional area.
Area factor =A1A2=3.0 mm21.5 mm2=2.0= \frac{A_1}{A_2} = \frac{3.0\text{ mm}^2}{1.5\text{ mm}^2} = 2.0.
Halving the cross-sectional area doubles the resistance due to inverse proportionality.
4
Calculate the final resistance of the replacement wire.
R2=0.85 Ω×0.60×2.0=1.02 ΩR_2 = 0.85\ \Omega \times 0.60 \times 2.0 = 1.02\ \Omega.
Combine the independent effects of length and cross-sectional area changes on the original resistance.

Anahtar Kavram

The electrical resistance of a conductor is directly proportional to its length and inversely proportional to its cross-sectional area (R=ρLAR = \rho \frac{L}{A}).
Tahmini Süre:1m 0s
Soru 174Soru →

An emergency egress lighting system operates from a 50 V50\text{ V} direct-current (DC) power source connected across two parallel branch circuits. Branch 1 contains an LED driver with an effective resistance of 25 Ω25\ \Omega. If the total current drawn from the power source is 3 A3\text{ A}, what is the resistance, in ohms (Ω\Omega), of Branch 2?

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Cevap: 50

Cevap

The resistance of Branch 2 is 50 Ω50\ \Omega.
In a parallel circuit, each branch receives the full source voltage of 50 V50\text{ V}. Branch 1 draws I1=50 V25 Ω=2 AI_1 = \frac{50\text{ V}}{25\ \Omega} = 2\text{ A}. Since the total current is 3 A3\text{ A}, Branch 2 must draw I2=3 A−2 A=1 AI_2 = 3\text{ A} - 2\text{ A} = 1\text{ A}. Applying Ohm's Law to Branch 2 gives R2=50 V1 A=50 ΩR_2 = \frac{50\text{ V}}{1\text{ A}} = 50\ \Omega.

Adım Adım Çözüm

1
Determine the voltage across each parallel branch.
V1=V2=50 VV_1 = V_2 = 50\text{ V}
In any parallel circuit configuration, the full source voltage appears equally across every individual parallel branch.
2
Calculate the current flowing through Branch 1 using Ohm's Law (I=VRI = \frac{V}{R}).
I1=50 V25 Ω=2 AI_1 = \frac{50\text{ V}}{25\ \Omega} = 2\text{ A}
Ohm's Law allows solving for branch current when branch voltage and resistance are known.
3
Calculate the remaining current flowing through Branch 2 by subtracting Branch 1 current from total line current.
I2=3 A−2 A=1 AI_2 = 3\text{ A} - 2\text{ A} = 1\text{ A}
According to Kirchhoff's Current Law, total current entering a parallel network must equal the sum of all individual branch currents (Itotal=I1+I2I_{\text{total}} = I_1 + I_2).
4
Calculate the resistance of Branch 2 using Ohm's Law (R=VIR = \frac{V}{I}).
R2=50 V1 A=50 ΩR_2 = \frac{50\text{ V}}{1\text{ A}} = 50\ \Omega
Dividing the branch voltage by the branch current yields the branch's effective resistance.

Anahtar Kavram

Parallel circuit branch voltage equality, Kirchhoff's Current Law (Itotal=I1+I2I_{\text{total}} = I_1 + I_2), and Ohm's Law (R=VIR = \frac{V}{I}).
Tahmini Süre:1m 0s
Soru 175Soru →

An electromagnetic security gate lock operates on an 18 V18\text{ V} direct-current (DC) power source. When fully energized, the lock draws a steady current of 250 mA250\text{ mA}. What is the electrical resistance of the lock in ohms (Ω\Omega)?

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Cevap: 72

Cevap

The electrical resistance of the electromagnetic lock is 72 Ω72\ \Omega.
According to Ohm's Law (V=I×RV = I \times R), resistance is found by dividing the potential difference by the current (R=VIR = \frac{V}{I}). Converting 250 mA250\text{ mA} to 0.25 A0.25\text{ A} and dividing 18 V18\text{ V} by 0.25 A0.25\text{ A} yields exactly 72 Ω72\ \Omega.

Adım Adım Çözüm

1
Convert the current measurement from milliamperes (mA\text{mA}) to amperes (A\text{A}).
I=250 mA=0.25 AI = 250\text{ mA} = 0.25\text{ A}
Ohm's Law calculations require current in base SI units (amperes) to yield resistance in ohms.
2
Rearrange Ohm's Law (V=I×RV = I \times R) to isolate resistance (RR).
R=VIR = \frac{V}{I}
Resistance is equal to the applied electromotive force (voltage) divided by the resulting current flow.
3
Substitute the given voltage and converted current into the formula and solve.
R=18 V0.25 A=72 ΩR = \frac{18\text{ V}}{0.25\text{ A}} = 72\ \Omega
Dividing 18 V18\text{ V} by 0.25 A0.25\text{ A} gives the total circuit resistance in ohms.

Anahtar Kavram

Ohm's Law states that resistance (RR) equals voltage (VV) divided by current (II), represented by the formula R=VIR = \frac{V}{I}.
Tahmini Süre:45s
Soru 176Soru →

A cylindrical water storage tank is filled to a depth of 4 meters4\text{ meters}, creating a hydrostatic pressure of 40,000 Pascals40,000\text{ Pascals} at the tank's flat bottom. If water is drained from the tank until the depth is exactly 1 meter1\text{ meter}, what is the new hydrostatic pressure at the bottom in Pascals? (Assume the density of water and gravity remain constant.)

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Cevap: 10000

Cevap

10000
Because hydrostatic pressure is directly proportional to depth (P=ρghP = \rho g h), reducing the depth to one-fourth of its original value (1 meter1\text{ meter} instead of 4 meters4\text{ meters}) also reduces the pressure exactly to one-fourth of its original value (40,000/4=10,00040,000 / 4 = 10,000).

Adım Adım Çözüm

1
Identify the relationship between hydrostatic pressure and fluid depth.
Hydrostatic pressure is directly proportional to depth (P∝hP \propto h) when density and gravity are constant.
This fundamental principle allows us to solve the problem using a simple ratio.
2
Calculate the pressure per unit of depth.
40,000 Pascals4 meters=10,000 Pascals per meter\frac{40,000\text{ Pascals}}{4\text{ meters}} = 10,000\text{ Pascals per meter}.
Finding the rate of pressure change per meter simplifies the calculation for the new depth.
3
Multiply the pressure per meter by the new depth.
10,000 Pascals/meter×1 meter=10,000 Pascals10,000\text{ Pascals/meter} \times 1\text{ meter} = 10,000\text{ Pascals}.
This yields the final pressure at the reduced depth.

Anahtar Kavram

Hydrostatic pressure is directly proportional to the depth of the fluid.
Soru 177Soru →

A specialized alloy heating wire has a length of 40 meters40\text{ meters}, a uniform cross-sectional area of 0.8 mm20.8\text{ mm}^2, and a material resistivity of 2.5×10−7 Ω⋅m2.5 \times 10^{-7}\ \Omega\cdot\text{m}. What is the total electrical resistance of this wire in ohms (Ω\Omega)?

Cevabı ve açıklamayı göster

Cevap: 12.5

Cevap

12.5 Ω
The electrical resistance of a uniform conductor is determined by R=ρLAR = \rho \frac{L}{A}. Converting 0.8 mm20.8\text{ mm}^2 to 8×10−7 m28 \times 10^{-7}\text{ m}^2 and substituting L=40 mL = 40\text{ m} along with ρ=2.5×10−7 Ω⋅m\rho = 2.5 \times 10^{-7}\ \Omega\cdot\text{m} yields (2.5×10−7)(40)8×10−7=12.5 Ω\frac{(2.5 \times 10^{-7})(40)}{8 \times 10^{-7}} = 12.5\ \Omega.

Adım Adım Çözüm

1
Convert the cross-sectional area from square millimeters to square meters.
A=0.8 mm2=0.8×10−6 m2=8×10−7 m2A = 0.8\text{ mm}^2 = 0.8 \times 10^{-6}\text{ m}^2 = 8 \times 10^{-7}\text{ m}^2
Resistivity is given in ohm-meters (Ω⋅m\Omega\cdot\text{m}), requiring all geometric dimensions to be in standard SI units (meters and square meters).
2
Apply the fundamental resistance formula relating resistivity, length, and cross-sectional area.
R=ρLAR = \rho \frac{L}{A}
The total resistance of a uniform conductor is directly proportional to its resistivity and length, and inversely proportional to its cross-sectional area.
3
Substitute the known values into the equation and solve for resistance.
R=(2.5×10−7 Ω⋅m)(40 m)8×10−7 m2=100×10−78×10−7=12.5 ΩR = \frac{(2.5 \times 10^{-7}\ \Omega\cdot\text{m})(40\text{ m})}{8 \times 10^{-7}\text{ m}^2} = \frac{100 \times 10^{-7}}{8 \times 10^{-7}} = 12.5\ \Omega
Canceling the common power factor 10−710^{-7} simplifies the expression to 100/8=12.5 Ω100 / 8 = 12.5\ \Omega.

Anahtar Kavram

Calculating conductor resistance from material resistivity, length, and cross-sectional area using R=ρL/AR = \rho L / A.
Tahmini Süre:1m 15s
Soru 178Soru →

A 24 V24\text{ V} direct-current (DC) power bus supplies two cooling fans connected in parallel inside a communications cabinet. The primary fan draws a steady branch current of 3 A3\text{ A}, while the secondary fan has an internal resistance of 12 Ω12\text{ }\Omega. What is the total current, in amperes (A\text{A}), delivered by the power bus to the parallel circuit?

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Cevap: 5

Cevap

The total current delivered by the power bus is 5 A5\text{ A}.
In a parallel DC circuit, the source voltage (24 V24\text{ V}) is applied directly across each branch. The secondary branch current is found using Ohm's law: I2=24 V12 Ω=2 AI_2 = \frac{24\text{ V}}{12\text{ }\Omega} = 2\text{ A}. Adding the known primary branch current (3 A3\text{ A}) to the secondary branch current (2 A2\text{ A}) yields the total source current of 5 A5\text{ A}.

Adım Adım Çözüm

1
Determine the voltage across the secondary branch
Voltage across the secondary fan is 24 V24\text{ V}
In any parallel circuit, every branch experiences the identical full source voltage (Vtotal=V1=V2V_{\text{total}} = V_1 = V_2).
2
Calculate the secondary branch current (I2I_2) using Ohm's law
I2=24 V12 Ω=2 AI_2 = \frac{24\text{ V}}{12\text{ }\Omega} = 2\text{ A}
Ohm's law states that current equals voltage divided by resistance (I=VRI = \frac{V}{R}).
3
Sum all individual branch currents to determine total circuit current
Itotal=3 A+2 A=5 AI_{\text{total}} = 3\text{ A} + 2\text{ A} = 5\text{ A}
Total current in a parallel circuit is the sum of the individual branch currents (Itotal=I1+I2I_{\text{total}} = I_1 + I_2).

Anahtar Kavram

Parallel Circuit Current and Ohm's Law
Tahmini Süre:1m 0s
Soru 179Soru →

An aircraft instrument panel illumination circuit is connected to a 28 V28\text{ V} direct-current (DC) power source. A multimeter measures a steady current flow of 350 mA350\text{ mA} through the circuit. What is the total electrical resistance of the circuit in ohms (Ω\Omega)?

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

The total electrical resistance of the circuit is 80 Ω80\ \Omega.
According to Ohm's Law, resistance equals voltage divided by current (R=VIR = \frac{V}{I}). Converting 350 mA350\text{ mA} to amperes yields 0.35 A0.35\text{ A}. Dividing 28 V28\text{ V} by 0.35 A0.35\text{ A} gives the correct resistance of 80 Ω80\ \Omega.

Adım Adım Çözüm

1
Convert the current from milliamperes to amperes
I=350 mA=0.35 AI = 350\text{ mA} = 0.35\text{ A}
Ohm's Law calculations require current to be expressed in standard amperes.
2
Apply Ohm's Law formula solved for resistance
R=VI=28 V0.35 A=80 ΩR = \frac{V}{I} = \frac{28\text{ V}}{0.35\text{ A}} = 80\ \Omega
Dividing the applied electromotive force by the current flow yields total circuit resistance.

Anahtar Kavram

Calculating resistance using Ohm's Law (R=VIR = \frac{V}{I}) with metric prefix conversion (1 A=1000 mA1\text{ A} = 1{}000\text{ mA}).
Tahmini Süre:45s
Soru 180Soru →

A quartermaster is organizing a drawer of 6060 uniform patches. The drawer contains 1515 rank insignia patches, 2121 unit patches, and the remaining patches are flag patches. If one patch is randomly selected from the drawer, what is the probability of selecting a flag patch? Express your answer as a decimal.

Cevabı ve açıklamayı göster

Cevap: 0.4

Cevap

0.4
The probability is found by dividing the number of flag patches (2424) by the total number of patches (6060). The fraction 2460\frac{24}{60} simplifies to 25\frac{2}{5}, which is 0.40.4 when expressed as a decimal.

Adım Adım Çözüm

1
Calculate the number of flag patches in the drawer.
60−(15+21)=60−36=2460 - (15 + 21) = 60 - 36 = 24 flag patches.
To determine the probability, the exact number of the desired item (flag patches) must first be known.
2
Set up the probability fraction.
2460\frac{24}{60}
The probability of a single event is the number of favorable outcomes divided by the total number of possible outcomes.
3
Convert the fraction to a decimal.
2460=25=0.4\frac{24}{60} = \frac{2}{5} = 0.4
The question explicitly specifies that the final answer must be expressed as a decimal.

Anahtar Kavram

Single Event Probability
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