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The network shown in the figure is a part of complete circuit. What is the potential difference `V_B-V_A` when the current `I` is `5 A` and is decreasing at a rate of `10^3A//s`? |
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Answer» In accordance with the law of potential distribution, for the given network, `V_(A) - IR + E - (dI)/(dt) L = V_(B)` And here `I` is decreasing (i.e., `dI//dt` is negative). `V_(B) - V_(A) = - 5 xx 1 + 15 - 5 xx 10^(-3)(-10^(3))` `V_(B) - V_(A) = - 5 + 15 + 5 = 15 V` |
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