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In the circuit shows in Fig. 6.63, the c...

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 galvanometer

B

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`

Text Solution

Verified by Experts

The correct Answer is:
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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