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In the circuit shown in figure, `E = 120 V, R_1 = 30.0Omega, R_2 = 50.0 Omega` and `L=0.200H`. Switch `S` is closed at `t = 0`. Just after the switch is closed. (a) What is the potential difference `V_(ab)` across the inductor `R_1`?(b) Which point, `a` or `b`, is at higher potential? (c) What is the potential difference `V_(cd)` across the inductor `L`? (d) Which point, `c` or `d`, is at a higher potential? The switch is left closed for a long time and then is opened. Just after the switch is opened (e) What is the potential difference `V_(ab)` across the resistor `R_1`? (f) Which point a or b, is at a higher potential? (g) What is the potential difference `V_(cd)` across the inductor `L`? (h) Which point, `c` or `d`, is at a higher potential? |
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Answer» Correct Answer - D These are two independent parallel circuits across the battery. a. `V_(ab)=E=120` volt (at all instants) b.`a` is higher potential. c. `V_(cd)` will decrease exponentially from `120V` to zero. `:. V_(cd)=120`volt, just after the switch is closed. d. `c` will be at higher potential. e. When switch is opeed, current through `R_1` will immediately become zero. While through `R_2`, will decrease to zero from the value `E/R_2=2.4A=i_0` (say), exponentially. Path of this decay of current will be `cdbac`. `:.` Just after the switch is opened. `V_(ab)=-i_0R_1=-2.4xx30=-72volt` f. Point `b` is at higher potential. g. `V_(cd)=-i_0(R_1+R_2)=-2.4(80)=-192`volt h. This time point `d` will be at higher potential. |
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