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InterviewSolution
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A capacitor of capacitance 5 µF is connected as shown in the figure. The internal resistance of the cell is 0.5 Ω. The amount of charge on the capacitor plates is(a) 80 µC (b) 40 µC (c) 20 µC (d) 10 µC |
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Answer» Answer is : (d) 10 µC In steady state, There will be no current in the capacitor branch. Net resistance of the first parallel branch and cell, R = 1+1+ 0.5 = 2.5Ω Current drawn from the cell, i = \(\frac{V}{R}\) = \(\frac{2.5}{2.5}\) = 1A Potential drop across two parallel branches, V = E - ir = 2.5 - 1\(\times\)0.5 = 2.5 - 0.5 2 0 V So, charge on the capacitor plates q = CV = 5\(\times\)2 = 10μC |
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