Question:
The value of current through the \(5\ \Omega\) resistor of the given circuit is
The value of current through the \(5\ \Omega\) resistor of the given circuit is
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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.
Updated On: Jun 25, 2026
- \(\dfrac{1}{25}\ \text{A}\)
- \(\dfrac{2}{25}\ \text{A}\)
- \(\dfrac{2}{23}\ \text{A}\)
- \(\dfrac{1}{23}\ \text{A}\)
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The Correct Option is D
Solution and Explanation
Step 1: Assign potentials to the circuit nodes.
Let the right junction be at zero potential: \[ V_R=0 \] From the \(5\ \text{V}\) battery, the left junction is at \[ V_L=5\ \text{V} \] Let the potential of the middle junction be \[ V_M=V \] In the upper branch, the \(3\ \text{V}\) cell makes the point after the \(5\ \Omega\) resistor at potential \[ 3\ \text{V} \]
Step 2: Apply Kirchhoff's current law at the middle junction.
Current through \(4\ \Omega\) resistor: \[ \frac{V-5}{4} \] Current through \(8\ \Omega\) resistor: \[ \frac{V-0}{8}=\frac{V}{8} \] Current through \(5\ \Omega\) resistor: \[ \frac{V-3}{5} \] Using Kirchhoff's current law, \[ \frac{V-5}{4}+\frac{V}{8}+\frac{V-3}{5}=0 \]
Step 3: Solve for \(V\).
Taking LCM \(40\), \[ 10(V-5)+5V+8(V-3)=0 \] \[ 10V-50+5V+8V-24=0 \] \[ 23V-74=0 \] \[ V=\frac{74}{23} \]
Step 4: Find current through \(5\ \Omega\) resistor.
Current through \(5\ \Omega\) resistor is \[ I=\frac{V-3}{5} \] Substituting \[ V=\frac{74}{23}, \] we get \[ I=\frac{\frac{74}{23}-3}{5} \] \[ I=\frac{\frac{74-69}{23}}{5} \] \[ I=\frac{5}{23}\times \frac{1}{5} \] \[ I=\frac{1}{23}\ \text{A} \]
Step 5: Final conclusion.
Hence, the current through the \(5\ \Omega\) resistor is \[ \boxed{\frac{1}{23}\ \text{A}} \]
Let the right junction be at zero potential: \[ V_R=0 \] From the \(5\ \text{V}\) battery, the left junction is at \[ V_L=5\ \text{V} \] Let the potential of the middle junction be \[ V_M=V \] In the upper branch, the \(3\ \text{V}\) cell makes the point after the \(5\ \Omega\) resistor at potential \[ 3\ \text{V} \]
Step 2: Apply Kirchhoff's current law at the middle junction.
Current through \(4\ \Omega\) resistor: \[ \frac{V-5}{4} \] Current through \(8\ \Omega\) resistor: \[ \frac{V-0}{8}=\frac{V}{8} \] Current through \(5\ \Omega\) resistor: \[ \frac{V-3}{5} \] Using Kirchhoff's current law, \[ \frac{V-5}{4}+\frac{V}{8}+\frac{V-3}{5}=0 \]
Step 3: Solve for \(V\).
Taking LCM \(40\), \[ 10(V-5)+5V+8(V-3)=0 \] \[ 10V-50+5V+8V-24=0 \] \[ 23V-74=0 \] \[ V=\frac{74}{23} \]
Step 4: Find current through \(5\ \Omega\) resistor.
Current through \(5\ \Omega\) resistor is \[ I=\frac{V-3}{5} \] Substituting \[ V=\frac{74}{23}, \] we get \[ I=\frac{\frac{74}{23}-3}{5} \] \[ I=\frac{\frac{74-69}{23}}{5} \] \[ I=\frac{5}{23}\times \frac{1}{5} \] \[ I=\frac{1}{23}\ \text{A} \]
Step 5: Final conclusion.
Hence, the current through the \(5\ \Omega\) resistor is \[ \boxed{\frac{1}{23}\ \text{A}} \]
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