TS EAMCET · Physics · Current Electricity
One end each of a resistance \(r\) capacitor \(C\) and resistance \(2 r\) are connected together. The other ends are respectively connected to the positive terminals of batteries, \(P, Q, R\) having respectively emf's \(E, E\) and \(2 E\). The negative terminals of the batteries are then connected together. In this circuit, with steady current the potential drop across the capacitor is :
- A \(\frac{E}{3}\)
- B \(\frac{E}{2}\)
- C \(\frac{2 E}{3}\)
- D \(E\)
Answer & Solution
Correct Answer
(A) \(\frac{E}{3}\)
Step-by-step Solution
Detailed explanation
In the steady state, no current flows through capacitor branch. Current in the circuit \(i=\frac{\text { net emf }}{\text { net resistance }}=\frac{2 E-E}{r+2 r}\) \(=\frac{E}{3 r}\) So, potential drop across capacitor \(V=i r=\frac{E}{3 r} \times r=\frac{E}{3}\)
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