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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 |
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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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