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What is the potential difference across 2μF capacitor in the circuit shown? A. 12 VB. 4 VC. 6 VD. 18 V

Answer» Correct Answer - C
Net emf in the circuit here,
`E=E_(2)-E_(1)=16-6=10V`
While the equivalent capacity
`C=(C_(1)C_(2))/(C_(1)+C_(2))=(2xx3)/(2+3)=(6)/(5) mu F`
Charge on each capacitor
`q=CV=(6)/(5) xx10 = 12 muC`
`therefore` Potential difference across 2 `muF` capacitor.
`V_(1)=(q)/(C_(1))=(12)/(2)=6V`


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