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)

D) e–Rt/LE) (1 – e–Rt/L)

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

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

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