Rate of change of current in an R-L Circuit

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Homework Statement


upload_2018-5-18_12-18-4.png


Homework Equations


Current cannot change suddenly through inductor unless input is impulse voltage source.

The Attempt at a Solution


since input is DC source we have i(0+) = i(0-) so we have di/dt at t = 0+ is 0.
Is it correct? Book has mentioned answer as B which is weird cause there is no Ls in circuit.
 

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  • #2
cnh1995
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Book has mentioned answer as B which is weird cause there is no Ls in circuit.
It's probably a typo. They meant Is instead of Ls.

since input is DC source we have i(0+) = i(0-) so we have di/dt at t = 0+ is 0.
No.
How about using source transformation? It will make the circuit look more familiar.
 
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  • #3
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Wow.... I used laplace and then got the answer. You're right.
upload_2018-5-18_16-41-37.png

But shouldn't it be zero as per concept? Unless the input is an impulse voltage source the current across inductor cannot change so quickly. V = L di/dt. So unless V is an impulse, di/dt has to be zero.
 

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  • #4
cnh1995
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Wow.... I used laplace and then got the answer. You're right.
Well, if you use source transformation, you end up with a series RL circuit with Rtotal=Rs+R and inductance L, excited by a voltage source of emf IsRs.
At t=0+, no current flows through the inductor, which means the entire source voltage appears across the inductor.
So, IsRs=Ldi/dt, which gives di/dt=IsRs/L.
the current across inductor cannot change so quickly. V = L di/dt. So unless V is an impulse, di/dt has to be zero.
No, here VL at t=0 is not zero, hence di/dt is also not zero. You can't compare the currents at t=0- and t=0+ to find di/dt at t=0+ because at t=0-, the inductor was not excited.
 
  • #5
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Well you're method is way quicker than mine.
Thanks a lot. Learnt new thing today.
 
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