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Find the charge on the four capacirtors of capacitances 1(mu)F,2(mu)F,3(mu)F,and 4(mu)F, shown in figure. |
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Answer» Solution :At steady state no current flows through the capacitors . ` R_eq = (3xx6)/(3+6) = 2 Omega ` ` i= 6/2 = 3 ` Since current is DIVIDED in the inverse ratio of resistance in each branch, 2` Omega ` will pass through ` 1, 2 Omega branch and one through 3, `3Omega` branch . ` V_AB = 2 xx 1 ` ` = 2V ` ` Q on 1 = MU f capacitors ` ` = 2 xx 1muc ` ` =2 MUC ` ` V_BC = 2 xx 2 = 4V. ` ` Q on 2 = 2 mu f capacitor = CV ` ` = 2 xx 4 mu c ` ` = 8 mu c ` ` V_DE = 3 xx 1 = 3V ` ` Q on 4 = 4 mu f capacitor ` ` = 3 xx 4 mu c ` ` =12 mu c ` ` V_FE = 3 xx 1 = 3V ` ` Q across 3 = 3 mu f capacitor ` ` 3 xx 3 mu c ` ` = 9 mu c ` . |
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