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A `muF` capacitor `C_1` is charged to `V_0 = 120 V`. The charging battery is then removed and the capacitor is connected in parallel to an uncharged `+ 4muF` capacitor `C_2`. (a) what is the potential difference `V` across the combination? (b) what is the stored energy before and after the switch `S` is closed? |
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Answer» Correct Answer - C::D a. `V=q_(n et)/C_(n et)=(C_1V_0)/(C_1+C_2)=V_0/(1+C_2//C_1)` `=120/(1+4//8)=80V` b. `U_i=1/2C_1V_0^2=1/2xx8xx10^-6xx(120)^2` `=5.76xx10^-2J` `U_f=1/2(C_1+C_2)V^2` `=1/2(C_1+C_2)V^2` `=1/2xx12xx10^-6xx(80)^2` `=3.81xx10^-2J` |
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