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In an `L-C` circuit, `L = 0.75 H` and `C =18 muF`, (a) At the instant when the current in the inductor is changing at a rate of `3.40 Ns`, what is the charge on the capacitor? (b) When the charge on the capacitor is `4.2 xx 10^-4` C, what is the induced emf in the inductor? |
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Answer» a. `V_L=V_C` `L(di)/(dt)=q/C` `q=(LC)=(di)/(dt)` `=0.75xx18xx10^Y-6)(3.4)` `=45.9xx10^-6C` `=45.9muC` b. `V_(L) = V_(C) = (q)/(C)` `= (4.2 xx 10^(-4))/(18 xx 10^(-6))` `= 23.3 V` |
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