Question

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

1. All voltmeters read 0 except $V_1$which reads $V_1=40V$and all ammeters read zero current except $A_1$and $A_4$read 0.8 A
2. role="math" localid="1668254558663" $V_1=24V,V_2=0and{V}_3=V_4=V_5=16V$.Also$A_1=0.48A,A_2=0.16A,A_3=0.32Aand{A}_4=0.$
3. The maximum charge on the capacitor is$Q=192\mu C$
4. The voltage increases exponentially to its maximum then decay exponentially to reach zero at$t=\infty$

Step-by-step solution

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.

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 voltmeters$V_{5}and{V}_3$reads zero

i.e$V_5={\mathrm{V}}_3=0$

Since the two voltmeters$V_{2}and{V}_3$ are connected in parallel with $V_4$, hence the summation of $V_2+V_3=V_4$.

We know $V_4=V_5=0$because they are connected in parallel and read the same voltage.

Since $V_4=V_5=V_3=0$we conclude that$V_2=0$ i.e. all voltmeters read zero voltage except which read the emf of the battery. Hence

$V_1=40V$

For the current,$A_1$ read current by

$\begin{array}{l}{I}_4=\frac{V_1}{R}\\ I_4=\frac{40V}{50\Omega }\\ I_1=0.8A\end{array}$

The current throughrole="math" localid="1668255083107" $A_{2}and{A}_3$ 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 $A_4$reads the same as$A_1$

role="math" localid="1668255109203" $A_1=A_4=08A$

Hence conclude that

All voltmeters read 0 except $V_1$which reads $V_1=40V$and all ammeters read zero current except $A_{1}and{A}_4$read 0.8 A

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

$\begin{array}{r}{R}_{eq}=\frac{\left(100\mathrm{\Omega }\right)\left(50\mathrm{\Omega }\right)}{(100\mathrm{\Omega }+50\mathrm{\Omega })}\\ R_{eq}=83.33\mathrm{\Omega }\end{array}$

The current though$A_1$ is

$\begin{array}{l}{I}_1=\frac{V_1}{R_{eq}}\\ I_I=\frac{40V}{83.33\Omega }\\ l_1=048A\end{array}$

The voltage across$V_1$ is calculated by Ohm’s law,

$\begin{array}{r}{V}_1=I_{1}R\\ V_1=\left(0.48\mathrm{A}\right)\left(50\mathrm{\Omega }\right)\\ V_1=24\mathrm{V}\end{array}$

As the inductor has zero voltage conclude $V_2$ reads 0 V.

The there voltmeters$V_3,V_{4}and{V}_5$ read the same voltage

$V_3=V_4=V_5=40V-24V=16V$

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

$\begin{array}{r}{I}_2=\frac{16\mathrm{V}}{100\mathrm{\Omega }}\\ I_2=0.16\mathrm{A}\\ I_3=\frac{32\mathrm{V}}{100\mathrm{\Omega }}\\ I_3=0.32\mathrm{A}\end{array}$

Hence We conclude that

role="math" localid="1668255488711" $\begin{array}{r}{V}_1=24V,V_2=0\text{and}{V}_3=V_4=V_5=16\mathrm{V}.\\ A_1=0.48A,A_2=0.16A,A_3=0.32\mathrm{A}\text{and}{A}_4=0.\end{array}$

Determine charge on capacitor

We have found that the voltage drop across the capacitor is$V_5$ and is equal todata-custom-editor="chemistry" $16\mathrm{V}$

The charge on a capacitor is given by

$\begin{array}{r}Q=CV\\ Q=\left(12\mu F\right)\left(16V\right)\\ Q=192\mu C\end{array}$

Curve of voltage across 

The voltage starts from zero and increases exponentially to it’s maximum and then decays exponentially to zero at$t=\infty$as shown in the diagram below: