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Two cells of emfs `E_1` and `E_2` and of negligible internal resistances are connected with two variable resistors as shown in Fig. A2.4. When the galvanometer shows no deflection, the values of the resistances are P and Q. What is the value of the ratio `E_2/E_1`? ` A. `P/Q`B. `P/(P+Q)`C. `Q/(P+Q)`D. `(P+Q)/P` |
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Answer» Correct Answer - B b. Potential difference `V_P` across P as determined from `E_1` id given by `V_P = (E_1/(P+Q))P.` Potential `V_P` across P as determined from `E_2` is same as `E_2` because no current is drawn, i.e., `V_P = E_2` Therefore, `E_2 = E_1(P/(P+Q)) or E_2//E_1 = (P/(P+Q))` . |
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