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WBJEE · Physics · Current Electricity

Four cells, each of emf \(E\) and internal resistance \(r,\) are connected in series across an external resistance \(R .\) By mistake one of the cells is connected in reverse. Then the current in the external circuit is

  1. A \(\frac{2 E}{4 r+R}\)
  2. B \(\frac{3 E}{4 r+R}\)
  3. C \(\frac{3 E}{3 r+R}\)
  4. D \(\frac{2 E}{3 r+R}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{2 E}{4 r+R}\)

Step-by-step Solution

Detailed explanation

Total emf of the cell \(=3 E-E=2 E\) Total internal resistance \(=4 r\) \(\therefore\) Total resistance of the circuit \(=4 r+R\) So, the current in the external circuit \(\left(\because i=\frac{V}{B}\right)\) \(\therefore\) \[ i=\frac{2 E}{4 r+R} \]
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