Question

In this circuit, the value of $I_2$ is

• A. 1.2A
• B. $I_1$ 10$\Omega$
• C. $I_2$ 15$\Omega$
• D. $I_3$ 30$\Omega$

Answer 1

0.4A

Answer 2

The circuit consists of three resistors $R_1 = 10\Omega$, $R_2 = 15\Omega$, and $R_3 = 30\Omega$ connected in parallel. The total current entering the parallel combination is $I = 1.2A$. We need to find the current $I_2$ flowing through the resistor $R_2 = 15\Omega$.

In a parallel circuit, the voltage across each branch is the same. Let $V$ be the voltage across the parallel combination. The current through each resistor is given by Ohm's law: $I_1 = \frac{V}{R_1}$, $I_2 = \frac{V}{R_2}$, $I_3 = \frac{V}{R_3}$

The total current is the sum of the currents in the parallel branches: $I = I_1 + I_2 + I_3 = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} = V \left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\right)$

We can calculate the equivalent resistance $R_{eq}$ of the parallel combination: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{10} + \frac{1}{15} + \frac{1}{30}$

The least common multiple of 10, 15, and 30 is 30. $\frac{1}{R_{eq}} = \frac{3}{30} + \frac{2}{30} + \frac{1}{30} = \frac{3+2+1}{30} = \frac{6}{30} = \frac{1}{5}$ So, $R_{eq} = 5\Omega$.

Now, we can find the voltage $V$ across the parallel combination using Ohm's law, $V = I \times R_{eq}$: $V = 1.2A \times 5\Omega = 6V$.

Now we can find the current $I_2$ through the resistor $R_2 = 15\Omega$: $I_2 = \frac{V}{R_2} = \frac{6V}{15\Omega} = \frac{6}{15}A = \frac{2}{5}A = 0.4A$.

Alternatively, we can use the current division rule. The ratio of currents in parallel branches is inversely proportional to the ratio of their resistances. Also, the current through a branch is the total current multiplied by the ratio of the equivalent resistance of the parallel combination to the resistance of that branch, or the total current multiplied by the ratio of the conductance of that branch to the total conductance.

Using the current division rule, for a parallel circuit with total current $I$ and branches with resistances $R_1, R_2, R_3$, the current through $R_2$ is given by: $I_2 = I \times \frac{\frac{1}{R_2}}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}} = I \times \frac{\frac{1}{R_2}}{\frac{1}{R_{eq}}} = I \times \frac{R_{eq}}{R_2}$ We calculated $R_{eq} = 5\Omega$. $I_2 = 1.2A \times \frac{5\Omega}{15\Omega} = 1.2A \times \frac{1}{3} = 0.4A$.

Both methods give the same result, $I_2 = 0.4A$.