InterviewSolution
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InterviewSolution
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In the circuit shows in Fig. 6.63, the cell is ideal with emf `9 V`. If the resistance of the coil of galvanometer is `1 Omega`, then A. no current flows in the galvanometerB. charge flowing through `8 mu F` is `40 mu C`C. potential difference across `10 mu F` is `5 V`D. potential difference across `10 mu F` is `4 V` |
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Answer» Correct Answer - A::B::D Since `R_(1)//R_(2) = C_(2)//C_(1)`, the wheatstone bridge is balanced. Hence, `V_(C) = V_(D)`. No current passes through the galvanometer. Hence, option (a) is correct. Potential difference across `R_(1) =` potential difference across `C_(1) = 4V` Potential difference across `R_(2) =` potential difference across `C_(2) = 5 V` Potential difference across `5 Omega` is the potential difference across `8 mu F` capacitor. So, charge across `8 mu F` capacitor is `Q = CV = 8 mu F xx 5V = 40 mu C`. hence, option (b) is correct and so is option(d). |
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