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When the current in the portion of the circuit shown in the figure is 2 A and increasing at the rate of `1As^(-1)`, then the measured potential difference `V_(ab)=8V`. However, when the current is 2 A and decreasing at the rate of `1As^(-1)`, then the measured potential difference `V_(ab)=4V`. The values of R and L are A. `3Omega` and 2 H respectivelyB. `3Omega` and 3 H respectivelyC. `2Omega` and 1 H respectivelyD. `3Omega` and 1 H respectively |
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Answer» Correct Answer - A From the figure, `V_(a)-iR-L(di)/(dt)=V_(b)` `therefore" "V_(a)-V_(b)=iR+L(di)/(dt)` According to given conditions, 8 = 2R + L ….(i) 4 = 2R - L …..(ii) Solving these two equation, we get `R=3Omega and L = 2H` |
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