Question

In the electrical circuit shown below, find the current in 4 $\Omega$ resistance.

Answer 1

4/3 A

Answer 2

Circuit Simplification and Current Division:

1. Parallel Combination: The two 4 $\Omega$ resistors at the bottom are in parallel. Their equivalent resistance is $R_{p1} = \frac{4 \Omega \times 4 \Omega}{4 \Omega + 4 \Omega} = 2 \Omega$.

2. Series Combination: The top 4 $\Omega$ resistor is in series with $R_{p1}$. This forms a branch (let's call it Branch 2) with a total resistance of $R_{Branch2} = 4 \Omega + 2 \Omega = 6 \Omega$.

3. Parallel Combination: The 3 $\Omega$ resistor (below the top 4 $\Omega$ resistor) is in parallel with Branch 2 (6 $\Omega$). Their equivalent resistance is $R_{p2} = \frac{3 \Omega \times 6 \Omega}{3 \Omega + 6 \Omega} = \frac{18}{9} = 2 \Omega$.

4. Total Series Resistance: The initial 3 $\Omega$ resistor (next to the battery) is in series with $R_{p2}$. The total equivalent resistance of the circuit is $R_{eq} = 3 \Omega + 2 \Omega = 5 \Omega$.

5. Total Current: The total current drawn from the 40 V source is $I_{total} = \frac{V}{R_{eq}} = \frac{40 V}{5 \Omega} = 8 A$.

6. Current Division: This $I_{total}$ current flows through the initial 3 $\Omega$ resistor. After this, it splits between the 3 $\Omega$ parallel resistor and Branch 2 (6 $\Omega$). The current flowing through Branch 2 is $I_{Branch2} = I_{total} \times \frac{3 \Omega}{3 \Omega + 6 \Omega} = 8 A \times \frac{3}{9} = 8 A \times \frac{1}{3} = \frac{8}{3} A$.

7. Current 'i': The current 'i' is indicated as flowing through one of the two 4 $\Omega$ resistors that are in parallel. Since these two resistors are equal and in parallel, the current $I_{Branch2}$ splits equally between them. Therefore, $i = \frac{I_{Branch2}}{2} = \frac{8/3 A}{2} = \frac{4}{3} A$.

Answer for Problem 1: The current in the 4 $\Omega$ resistance (referring to one of the bottom parallel 4 $\Omega$ resistors) is $\frac{4}{3} A$.