Question

An inductance L, resistance R, and ideal battery of emf are wired in series. A switch in the circuit is closed at time t = 0, at which time the current is zero. At any later time t the emf of the inductor is given by:The question is aksing the emf of inducotr not resistor, the answer is D. Could anyone explain it to me? thank youA) (1 – e–Lt/R)B) e–Lt/RC) (1 + e–Rt/L)D) e–Rt/LE) (1 – e–Rt/L)

Answer 1

D) e–Rt/LE) (1 – e–Rt/L)

Step-by-step explanation:

After some time t the current does not passing through the circuit

=>so the back emf is zero

=>here the inductor opposes decay of the circuit

-Ldi/dt=Ri

di/dt=-R/Li

di/i=-R/Ldt

now we applying the integration on both sides

log i=-R/Lt+C

here t=0=>i=io

Log io=C

=>Log i=-R/L×t+Log io

logi-Log io=-R/L×t

Log[i/io]=-R/L*t

i/io=e^-Rt/L

i=ioe^-Rt/L

the option D is correc

Answer 2

Step-by-step explanation:

After some time t the current does not passing through the circuit

=>so the back emf is zero

=>here the inductor opposes decay of the circuit

- Ldi/dt = Ri

di/dt = - R/Li

di/i = - R/Ldt

now we applying the integration on both sides

log i=-R/Lt+C

here t=0=>i=io

Log io=C

=>Log i=-R/L*t + Log io

logi-Log io=-R/L*t

Log[i/io]=-R/L*t

i/io=e^-Rt/L

i=ioe^-Rt/L

the option D is correct