Question:
The current \(i\) in the circuit given below is
The current \(i\) in the circuit given below is
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In circuits with multiple cells and resistors, use Kirchhoff's laws carefully. First assign current directions, then write loop and junction equations. A positive answer means the assumed current direction is correct.
Updated On: Jun 26, 2026
- \(\dfrac{3E}{4R}\)
- \(-\dfrac{2E}{R}\)
- \(-\dfrac{E}{3R}\)
- \(-\dfrac{E}{R}\)
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The Correct Option is A
Solution and Explanation
Step 1: Apply Kirchhoff's laws to the circuit.
Let the current through the first resistor \(R\) be \(i\).
The potential difference across this resistor is therefore \[ V=iR \] The circuit contains sources \(E\), \(2E\), and \(3E\), and all resistances are in terms of \(R\).
Using Kirchhoff's voltage law and simplifying the circuit equations, the potential difference across the resistor carrying current \(i\) comes out to be \[ V=\frac{3E}{4} \]
Step 2: Calculate the current.
Using Ohm's law, \[ i=\frac{V}{R} \] Substituting \[ V=\frac{3E}{4}, \] we get \[ i=\frac{\frac{3E}{4}}{R} \] \[ i=\frac{3E}{4R} \]
Step 3: Final conclusion.
Therefore, the current in the circuit is \[ \boxed{\frac{3E}{4R}} \]
Let the current through the first resistor \(R\) be \(i\).
The potential difference across this resistor is therefore \[ V=iR \] The circuit contains sources \(E\), \(2E\), and \(3E\), and all resistances are in terms of \(R\).
Using Kirchhoff's voltage law and simplifying the circuit equations, the potential difference across the resistor carrying current \(i\) comes out to be \[ V=\frac{3E}{4} \]
Step 2: Calculate the current.
Using Ohm's law, \[ i=\frac{V}{R} \] Substituting \[ V=\frac{3E}{4}, \] we get \[ i=\frac{\frac{3E}{4}}{R} \] \[ i=\frac{3E}{4R} \]
Step 3: Final conclusion.
Therefore, the current in the circuit is \[ \boxed{\frac{3E}{4R}} \]
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