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A `4muF` capacitor is charged to 400 V. If its plates are joined through a resistance of `2kOmega`, then heat produced in the resistance isA. 0.16 JB. 0.32 JC. 0.64 JD. 1.28 J |
Answer» Correct Answer - B `C=4muF,V=400V,R=2kOmega,H=?` `H=E=1/2CV^(2)` `=(4xx10^(-6)xx16xx10^(4))/2=64/2xx10^(-2)=0.32J` |
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