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Determine the potential difference across the plates of the capacitorC_1 of the network shown in the Fig. Assume epsi_1 gt epsi_2. |
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Answer» Solution :As `epsi_1 gt epsi_2` HENCE the net effective voltage PRESENT in the network `EPSI= epsi_1 - epsi_2`. In the network `C_1 and C_2` are joined in series, hence EQUIVALENT capacitance of the network `C = (C_1C_2)/(C_1 + C_2)`. `:.`Charge one each capacitor q `= C epsi =((C_1C_2)/(C_1 + C_2))(epsi_1 - epsi_2)` `:.` Potential DIFFERENCE across the plate of the capacitor `C_1`, is `V = q/(C_1) = (C_2/(C_1 + C_2))(epsi_1 - epsi_2)` |
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