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The figure shows a graph of the current in a charging circuit of a capacitor through a resistor of resistance `10Omega`. A. the initial potential difference across the capacitor is `100V`B. The capacitor of the capcitor is `1/(10ln2) F`C. the total heat produced in the circuit will be `(500/ln2)J`D. All of the above |
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Answer» Correct Answer - D `V_0=i_0R=(10)(10)=100V` after `2s` , current becoms `1/4` th. Therefore, after `1s`, current will remain half also called half life. `t_(1/2)=(ln2)tau_C=(ln 2)CR` `:. C=((t_(1/2)))/((ln 2)R)=1/(10 ln 2) F` total heat `=1/2 CV_0^2` `=1/2x1/(10 ln 2) (100)^2` `=500/(ln 2) J` |
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