Question:
The current \(i\) in the circuit shown in the figure is
The current \(i\) in the circuit shown in the figure is
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In complex resistor networks, first reduce parallel and series combinations, then apply Kirchhoff's voltage law carefully with correct emf polarities.
Updated On: Jun 22, 2026
- \(\dfrac{\varepsilon}{R}\)
- \(-\dfrac{\varepsilon}{R}\)
- \(\dfrac{2\varepsilon}{R}\)
- \(-\dfrac{2\varepsilon}{R}\)
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The Correct Option is A
Solution and Explanation
Step 1: Simplify the right-side parallel resistors.
On the right side of the circuit, two resistors \(R\) are connected in parallel.
Equivalent resistance is \[ R_p=\frac{R\cdot R}{R+R} \] \[ R_p=\frac{R}{2} \] This equivalent resistance is connected in series with the bottom resistor \(R\).
Hence, effective resistance of the right loop is \[ R+\frac{R}{2} = \frac{3R}{2} \]
Step 2: Apply Kirchhoff's loop law.
Let the current through the top resistor be \(i\) in the indicated direction.
Traversing the upper loop and applying KVL, the net emf in the loop is obtained from the sources: \[ 3\varepsilon-2\varepsilon-\varepsilon=0 \] Thus the remaining voltage drop occurs only across the resistor carrying current \(i\).
Hence, \[ iR=\varepsilon \]
Step 3: Find the current.
\[ i=\frac{\varepsilon}{R} \]
Step 4: Final conclusion.
Therefore, the current in the circuit is \[ \boxed{\frac{\varepsilon}{R}} \]
On the right side of the circuit, two resistors \(R\) are connected in parallel.
Equivalent resistance is \[ R_p=\frac{R\cdot R}{R+R} \] \[ R_p=\frac{R}{2} \] This equivalent resistance is connected in series with the bottom resistor \(R\).
Hence, effective resistance of the right loop is \[ R+\frac{R}{2} = \frac{3R}{2} \]
Step 2: Apply Kirchhoff's loop law.
Let the current through the top resistor be \(i\) in the indicated direction.
Traversing the upper loop and applying KVL, the net emf in the loop is obtained from the sources: \[ 3\varepsilon-2\varepsilon-\varepsilon=0 \] Thus the remaining voltage drop occurs only across the resistor carrying current \(i\).
Hence, \[ iR=\varepsilon \]
Step 3: Find the current.
\[ i=\frac{\varepsilon}{R} \]
Step 4: Final conclusion.
Therefore, the current in the circuit is \[ \boxed{\frac{\varepsilon}{R}} \]
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