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Zorluk: KolaySeries Circuits

A 24 V24\text{ V} direct current (DC) power supply is connected to a circuit with two resistors in series: a 4 Ω4\ \Omega resistor and an 8 Ω8\ \Omega resistor. What is the current flowing through the 8 Ω8\ \Omega resistor?

  1. A
    3 A3\text{ A}
  2. 2 A2\text{ A}Cevap
  3. C
    6 A6\text{ A}
  4. D
    9 A9\text{ A}

Cevap

2 A2\text{ A}
The total resistance of the series circuit is the direct sum of the resistors (4 Ω+8 Ω=12 Ω4\ \Omega + 8\ \Omega = 12\ \Omega). According to Ohm's Law (I=V/RI = V / R), dividing the 24 V24\text{ V} source by 12 Ω12\ \Omega yields a total current of 2 A2\text{ A}. Because a series circuit provides only a single continuous path for electron flow, the current is identical through every component in the loop, meaning 2 A2\text{ A} flows through the 8 Ω8\ \Omega resistor.

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1
Calculate the total equivalent resistance of the series circuit.
Rtotal=R1+R2=4 Ω+8 Ω=12 ΩR_{\text{total}} = R_1 + R_2 = 4\ \Omega + 8\ \Omega = 12\ \Omega
In any series circuit, individual resistances add directly together to form the total circuit resistance.
2
Apply Ohm's Law to determine the total circuit current.
Itotal=VtotalRtotal=24 V12 Ω=2 AI_{\text{total}} = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{24\text{ V}}{12\ \Omega} = 2\text{ A}
Current equals total voltage divided by total resistance.
3
Determine the current flowing specifically through the 8 Ω8\ \Omega resistor.
IR2=Itotal=2 AI_{R2} = I_{\text{total}} = 2\text{ A}
In a single-loop series circuit, current is uniform at all points and through all components.

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Series Circuit Current and Total Resistance
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Series Circuits

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