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Zorluk: OrtaParallel Circuits

A direct-current power supply in an automotive workshop delivers 24 V24\text{ V} across two parallel branches. The first branch contains an 8 Ω8\ \Omega inspection lamp, and the second branch contains a 24 Ω24\ \Omega ventilation motor. What is the total current drawn by the circuit from the power supply?

  1. A
    0.75 A0.75\text{ A}
  2. 4 A4\text{ A}Cevap
  3. C
    3 A3\text{ A}
  4. D
    6 A6\text{ A}

Cevap

4 A4\text{ A}
In a parallel circuit, the voltage across each branch is identical to the supply voltage (24 V24\text{ V}). The current in each branch is calculated using Ohm's Law: the 8 Ω8\ \Omega load draws 3 A3\text{ A} (24 V/8 Ω24\text{ V} / 8\ \Omega), and the 24 Ω24\ \Omega load draws 1 A1\text{ A} (24 V/24 Ω24\text{ V} / 24\ \Omega). Summing these branch currents gives a total supply current of 4 A4\text{ A}.

Adım Adım Çözüm

1
Calculate the current flowing through the first branch (8 Ω8\ \Omega resistor).
I1=VR1=24 V8 Ω=3 AI_1 = \frac{V}{R_1} = \frac{24\text{ V}}{8\ \Omega} = 3\text{ A}
In a parallel circuit, each branch receives the full source voltage of 24 V24\text{ V}.
2
Calculate the current flowing through the second branch (24 Ω24\ \Omega resistor).
I2=VR2=24 V24 Ω=1 AI_2 = \frac{V}{R_2} = \frac{24\text{ V}}{24\ \Omega} = 1\text{ A}
Ohm's Law applies independently to the second branch using the same 24 V24\text{ V} potential.
3
Sum the individual branch currents to find the total circuit current.
Itotal=I1+I2=3 A+1 A=4 AI_{\text{total}} = I_1 + I_2 = 3\text{ A} + 1\text{ A} = 4\text{ A}
According to Kirchhoff's Current Law, the total current supplied to a parallel network equals the sum of all branch currents.

Anahtar Kavram

In a parallel circuit, total current is the sum of the individual branch currents, each determined by dividing the constant source voltage by the branch resistance.

Alternatif Yöntem

Find the equivalent parallel resistance first: Req=R1×R2R1+R2=8×248+24=19232=6 ΩR_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{8 \times 24}{8 + 24} = \frac{192}{32} = 6\ \Omega. Then apply Ohm's Law to the whole circuit: Itotal=VReq=24 V6 Ω=4 AI_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{24\text{ V}}{6\ \Omega} = 4\text{ A}.
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