A 3-$\Omega$and a 1.5-$\Omega$ resistor are wired in parallel and the combination is wired in series to a 4-$\Omega$ resistor and a 10-V emf device.The current in the 3-$\Omega$ resistor is:
A)0.33 A
B)0.67 A
C)2.0 A
D)3.3 A
E)6.7 A

Question

Quizplus · Accepted Answer

First, calculate the equivalent resistance of the 3-ohm and 1.5-ohm resistors in parallel: $R_{eq} = \frac{1}{\frac{1}{3} + \frac{1}{1.5}} = 1 \Omega$. Then, add this to the 4-ohm resistor in series for a total resistance of $5 \Omega$. The total current from the 10-V source is $I = \frac{V}{R} = \frac{10V}{5\Omega} = 2A$. Since the resistors in parallel have the same voltage across them, the current through the 3-ohm resistor is $I = \frac{V}{R} = \frac{10V}{3\Omega} = \frac{10}{3}A \approx 3.33A$, but this calculation mistake leads to reconsideration. Correctly, the total current (2A) is divided among the parallel resistors inversely by their resistance values. The voltage across the parallel combination is $V = I_{total} \times R_{eq} = 2A \times 1\Omega = 2V$. Thus, the current through the 3-ohm resistor, using the parallel voltage, is $I = \frac{V}{R} = \frac{2V}{3\Omega} \approx 0.67A$, which corrects the initial mistake in the explanation.

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