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A capacitors `C_1` is charged to a potential `V` and connected to another capacitor in seris with a resistor `R` as shown. It is observed that heat `H_1` is dissipated across resistance `R`, tilll the circuit reaches steady state. Same process is repeated using resistance of `2R`. If `H_2` is heat dissipated in this case then A. `H_2/H_1=1`B. `H_2/H_1=4`C. `H_2/H_1=1/4`D. `H_2/H_1=2` |
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Answer» Correct Answer - A `H_1=H_2=U_i-U_f` the only change is by increasing the resistance `tau_C` increase. Hence process of radistribution of charge slows down. |
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