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Two capacitors `C_(1)` and `C_(2)` are connected in a circuit as shown in figure. The potential difference `(V_(A) - V_(B))` is A. `8V`B. `-8V`C. `12V`D. `-12V` |
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Answer» Correct Answer - B By using KVL A to A `12 + q/4 - 24 + q/2 = 0` `q/2 + q/4 = 12` `q = (12 xx 4)/(3) = 16 muC` `V_(A) - V_(B) = (-q)/(2muC) = - 8V` |
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