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In the circuit cell are of equal `emf E` but of difference internal resistance `r_(1) = 6 Omega` reading of the ideal voltmeter connected across cell `1` is zero The value of the external resistance `R` is ohm is equal to A. `10 Omega`B. `4 Omega`C. `2 Omega`D. `2.4 Omega` |
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Answer» Correct Answer - c Current in the circuit `1 = (E+E)/(r_(1) + (r_(2) + R)) = (2E)/(6+4+R) = (2E)/(10+R)` Potential difference across `(R+ r_(2))` `= (2E)/((10+R)) (R+r_(2))` As voltmeter connecting cell `1` records zero reading therefore the e.m.f of cell `2` is equal to voltage drop across `(R+r_(2))` hence `E = (2E)/((10+R))(R+r_(2))= (2E(R+4))/(10+R)` On solving `R = 2 Omega` |
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