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In the circuit shown, when switch `S_(1)` is closed and `S_(2)` is open, the ideal voltmeter shows a reaiding of `18 V`. When switch `S_(2)` is closed and `S_(1)` is open, the reading of voltmeter is `24 V`. When `S_(1)` and `S_(2)` both are closed, the voltmeter reading will be A. `14.4 V`B. `20.6 V`C. `24.4 V`D. `10.8 V` |
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Answer» Correct Answer - A (a) Terminal potential difference across the terminals of cell `E` `V = E - ir = E - ((E)/(r + R)) r = (ER)/(R + r)` According to question `18 = (6 E)/(6 + r)` and `24 = (12 E)/(12 + r)` Solving (i) and (ii), we get `r = 6 Omega, E = 36 V` When both `S_(1)` and `S_(2)` are closed, the effective external resistance in the circuit will be. `R = (6 xx 12)/(18) = 4 Omega` So, `V = (ER)/(r + R) = (36 xx 4)/(6 + 4) = (144)/(10) = 144` volt `V = 14.4` volt Thus, choice (a) is correct. |
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