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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
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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)
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Physics
4 years
2021-08-02T15:25:32+00:00
2021-08-02T15:25:32+00:00 2 Answers
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Answers ( )
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
Answer:
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