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In the circuit shown in Fig. P30.59, switch Sis closed at time t= 0 with no charge initially on the capacitor.

(a) Find the reading of each ammeter and each voltmeter just after Sis closed.

(b) Find the reading of each meter after a long time has elapsed.

(c) Find the maximum charge on the capacitor.

(d) Draw a qualitative graph of the reading of voltmeter V2 as a function

of time.

Short Answer

Expert verified
  1. All voltmeters read 0 except V1which reads V1=40Vand all ammeters read zero current except A1and A4read 0.8 A
  2. role="math" localid="1668254558663" V1=24V,V2=0andV3=V4=V5=16V.AlsoA1=0.48A,A2=0.16A,A3=0.32AandA4=0.
  3. The maximum charge on the capacitor isQ=192μC
  4. The voltage increases exponentially to its maximum then decay exponentially to reach zero att=

Step by step solution

01

Important Concepts

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 = lR

An inductor act as a wire of infinite resistance right after the circuit is closed while it acts as a normal conducting wire after a long time of circuit being closed

A capacitor acts as normal conducting wire when the circuit is just closed while it acts a wire of infinite resistance after a long time of circuit being closed.

Same current flows through the ammeters connected in series.

Same voltage is dropped across voltmeters connected in parallel.

02

Determine the voltages and Current in Part A

In the case just after the switch is close, and inductor has zero current flowing through it and it is considered to be a wire of infinite resistance i.e. the circuit is considered open at this branch meanwhile the voltage across any capacitor is zero and it is considered to be a conducting wire.

Hence we determine that the voltage drop across voltmetersV5andV3reads zero

i.eV5=V3=0

Since the two voltmetersV2andV3 are connected in parallel with V4, hence the summation of V2+V3=V4.

We know V4=V5=0because they are connected in parallel and read the same voltage.

Since V4=V5=V3=0we conclude thatV2=0 i.e. all voltmeters read zero voltage except which read the emf of the battery. Hence

V1=40V

For the current,A1 read current by

I4=V1RI4=40V50ΩI1=0.8A

The current throughrole="math" localid="1668255083107" A2andA3 is zero as they read the current through the inductor and capacitor respectively.

As the capacitor acts as a conducting wire, the current thought the ammeter A4reads the same asA1

role="math" localid="1668255109203" A1=A4=08A

Hence conclude that

All voltmeters read 0 except V1which reads V1=40Vand all ammeters read zero current except A1andA4read 0.8 A

03

Determine the voltages and Current in Part B

In the second case of after a long time , any inductor is considered as a conducting wire hence has zero voltage drop across it. Current through the capacitor is zero and its resistance is considered infinite i.e. the circuit is considered open at this branch.

We need to calculate the equivalent resistance

Req=(100Ω)(50Ω)(100Ω+50Ω)Req=83.33Ω

The current thoughA1 is

I1=V1ReqII=40V83.33Ωl1=048A

The voltage acrossV1 is calculated by Ohm’s law,

V1=I1RV1=(0.48A)(50Ω)V1=24V

As the inductor has zero voltage conclude V2 reads 0 V.

The there voltmetersV3,V4andV5 read the same voltage

V3=V4=V5=40V-24V=16V

Now find the current through each ammeter using Ohm’s law

I2=16V100ΩI2=0.16AI3=32V100ΩI3=0.32A

Hence We conclude that

role="math" localid="1668255488711" V1=24V,V2=0andV3=V4=V5=16V.A1=0.48A,A2=0.16A,A3=0.32AandA4=0.

04

Determine charge on capacitor

We have found that the voltage drop across the capacitor isV5 and is equal todata-custom-editor="chemistry" 16V

The charge on a capacitor is given by

Q=CVQ=(12μF)(16V)Q=192μC

05

Curve of voltage across 

The voltage starts from zero and increases exponentially to it’s maximum and then decays exponentially to zero att=as shown in the diagram below:

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