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Drupady [299]
3 years ago
13

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)
Physics
2 answers:
Kay [80]3 years ago
3 0

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

4vir4ik [10]3 years ago
3 0

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

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Answer:

Explanation:

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r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.

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b)

Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre

9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]

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1/2 m v² = 12 x 10⁻⁵ J

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Answer:

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