# 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.