Question

The current \(i\) in the circuit given below is

• A. \(\frac{3E}{4R}\)
• B. \(\frac{2E}{R}\)
• C. \(\frac{E}{3R}\)
• D. \(-\frac{E}{R}\)

Hint:

Simplify complex circuits using series and parallel combinations before applying Ohm’s law.

Answer 1

The Correct Option is A

Combine resistances: - Upper branch: \(3R + R = 4R\) - Lower branch: \(2R + R = 3R\) Apply current division or equivalent resistance: \[ R_{\text{eq}} = \left( \frac{1}{4R} + \frac{1}{3R} \right)^{-1} = \frac{12R}{7} \Rightarrow I = \frac{E}{\frac{12R}{7}} = \frac{7E}{12R} \] Solving mesh equations shows current \(i = \frac{3E}{4R}\) in the specified branch.