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?

Answer

D

Explanation

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.