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In the circuit shown in figure, the capacitor is charged with a cell of `5V`.If the switch is closed at `t=0`, then at `t=12s`, charge on the capacitor is A. `(0.37)10muC`B. `(0.37)^210muC`C. `(0.63)10muC`D. `(0.63)^210muC` |
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Answer» Correct Answer - B `tau_c=CR=6s` `q_0=CV=10muC` `Now, q=q_0e^(-t/tauC)=(10muC)e^(-12/6)` `=(1/6)^2(10muC)` `=(0.37)^2(10muC)` |
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