GRE GR 1777, Problem 004 Solution

Physics GRE GR 1777
4. Electrodynamics
Solution) Net resistance in a parallel circuit is
\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{1} + \frac{1}{2} = \frac{3}{2}
\therefore R_{parallel} = \frac{2}{3}
Then, total resistance in circuit is
R_{total} = 1 + \frac{2}{3} = \frac{5}{3}.
By the Ohm’s law, V = IR, the total current is 20 = \frac{5}{3} \cdot I,
\therefore I = 12A.
In a parallel circuit, the current is divided. The ratio of the individual current is the inverse ratio as individual resistance, so
R_{1} : R_{2} = 1 : 2 = 1/I_{1} : 1/I_{2}.
Therefore, the current i through the 2\Omega resistor is 4A.
The answer is (B).

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