Question

The value of current through the \(5\ \Omega\) resistor of the given circuit is

• A. \(\dfrac{1}{25}\ \text{A}\)
• B. \(\dfrac{2}{25}\ \text{A}\)
• C. \(\dfrac{2}{23}\ \text{A}\)
• D. \(\dfrac{1}{23}\ \text{A}\)

Hint:

For circuits with multiple cells and resistors, assign node potentials and apply Kirchhoff's current law at a junction. This method avoids confusion with current directions.

Answer 1

The Correct Option is D

Step 1: Label nodes and assign potentials.
Reference: right node at $V_R = 0$ V. The 5 V battery places the left node at $V_L = 5$ V. Let the middle junction potential be $V$. The 3 V cell places its upper branch node at $3$ V.
Step 2: Write branch currents using Ohm's law.
Current through $4\,\Omega$ (left to middle): $I_4 = \frac{5-V}{4}$. Current through $8\,\Omega$ (middle to right): $I_8 = \frac{V}{8}$. Current through $5\,\Omega$ (from upper 3 V node toward middle): $I_5 = \frac{V-3}{5}$.
Step 3: Apply KCL at the middle junction.
\[ \frac{5-V}{4} - \frac{V}{8} - \frac{V-3}{5} = 0 \]
Step 4: Solve for $V$ (multiply through by LCM = 40).
\[ 10(5-V) - 5V - 8(V-3) = 0 \Rightarrow 50-10V-5V-8V+24=0 \Rightarrow 74=23V \Rightarrow V = \frac{74}{23} \text{ V} \]
Step 5: Find the current through the $5\,\Omega$ resistor.
\[ I_5 = \frac{V-3}{5} = \frac{\frac{74}{23}-3}{5} = \frac{5/23}{5} = \frac{1}{23} \text{ A} \]
Step 6: State the final answer.
\[ \boxed{I_{5\,\Omega} = \frac{1}{23} \text{ A}} \]