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Consider the circuit shown in Fig. P30.63.Let E = 36.0 V, R0 = 50.0 Ω, R = 150 Ω, and L = 4.00 H.
(a) Switch S1 is closed and switch S2 is left open. Just after S1 is closed, what are the current i0 through R0 and the potential differences vac and vcb ? (b) After S1 has been closed a long time (S2 is still open) so that the current has reached its final, steady value, what are i0 , Vac , and Vcb ? (c) Find the expressions for i0 , Vac , and Vcb as functions of the time t since S1 was closed. Your results should agree with part (a) when t = 0 and with part (b) whent. Graph i0 , Vac ,and Vcb versus time.

Short Answer

Expert verified
  1. The voltmeters readVab=0,Vbc=36V and the current isi0=0
  2. The voltmeters readVab=9,Vbc=27V and the current is i0=0.18A.
  3. The current isi0=0.180A(1-e-t0.02s)

Step by step solution

01

Important Concepts and Formula

Ohm’s law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant

V = IR

According to Kirchhoff’s Lawthe sum of the voltages around the closed loop is equal to null.

V=0

02

Voltage Drop when switch is just closed

When the switch is just closed the initial current is zero i.ei0=0A and in this case the voltage across ab will be zero Vab=0.

And as no voltage drop occurs therefore the voltage across bc is the same of the emf of the batteryVbc=36V

Hence the voltmeters readVab=0,Vbc=36V and the current is i0=0.

03

Voltage Drop in steady state

After a long time of closing of the switch, the voltage across the inductor is zero, so use the ohm’s law to geti0.

i0=εReqi0=36V(150Ω+30Ω)i0=0.18A

Now use this value to get the voltage acrossVab andVbc by

Vab=i0R0=(0.18A)(50Ω)=9V

And

Vbc=i0R=(0.18A)(150Ω)=27V

Hence The voltmeters readVab=9,Vbc=27V and the current is i0=0.18A.

04

Using Kirchhoff’s Law

Use Kirchhoff’s rule in the bottom loop and get the equation

εVacVcb=0εi0R0i0RLdidt=0

Rearranging

Ldidt=εi0R0+Rdii+εR0+R=R0+RLdt

Integrate both sides

lni+εR0+RεR0+R=R0+RLdt

Take exponent on both sides

i+εR0+RεR0+R=eR0+RLdt

Rearrange to get

i0=εR0+R1eR0+RLt

Plug in the valuesR,R0,Landε of to get

i0=0.180A1et0.02s

Hence the current isi0=0.180A1et0.02s

05

Plot Graphs

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