# Electric Circuits Questions

### Questions

The circuit shown in the figure, has reached equilibrium (steady state) and the power output of the $2\Omega$ resistor is $8\ {\rm W}$. The switch is closed, and the circuit is once again given enough time to come to equilibrium. Find the change in the charge on each capacitor from the initial equilibrium state to the final equilibrium state.

In the following circuit, the capacitors are initially uncharged. At $t=0$, the switch is closed. Find the following at $10\ {\rm ms}$. (${\mathcal E}=12{\rm V}$)
(a) The charge on each capacitor
(b) The current through each resistor

(a) First after closing the switch, find the equivalent resistance and capacitance to be able to determine the time constant of the circuit i.e. $\tau=RC$.
The two resistors and capacitors are in parallel so $C_{eq}=6+3=9\ \mu {\rm F}$ and $R_{eq}=\frac{6\times 3}{6+3}=2\ {\rm k}\Omega$ thus $\tau=R_{eq}C_{eq}=2\times 9=18\ {\rm ms}$
We know that in charging mode the charge on the capacitor at any moment of time is given by $Q\left(t\right)=C{\mathcal E}\left(1-e^{-\frac{t}{\tau}}\right)$. Therefore, the total charge on the capacitors after $t=10\ {\rm ms}$ is $Q\left(10\ {\rm ms}\right)=C_{eq}{\mathcal E}\left(1-e^{-\frac{t}{\tau}}\right)=\left(9\times 12\right)\left(1-e^{-\frac{10}{18}}\right)=46\ \mu {\rm C}$
Now with the given $Q$, we can find the potential difference $\Delta V_{ab}$ across the capacitors
$\Delta V_{ab}=\frac{Q_{10\ {\rm ms}}}{C_{eq}}=\frac{46}{9}=5.115\ {\rm V}$
All components in parallel have the same potential drop i.e. $\Delta V_{6\mu F}=\Delta V_{3\mu F}$
$\Rightarrow Q_6=C_6\Delta V_6=6\times 5.115=30.7\ \mu {\rm C}$
$Q_3=C_3\Delta V_3=3\times 5.115=15.3\ \mu {\rm C}$
(b) In charging mode the current through the circuit is given by $I\left(t\right)=\frac{{\mathcal E}}{R_{eq}}{\exp \left(-\frac{t}{\tau}\right)\ }$. Therefore, $I\left(t=10\ {\rm ms}\right){\rm =}\frac{{\rm 12}}{{\rm 2000}}{\exp \left(-\frac{10}{18}\right)\ }=3.44\ {\rm mA}$. The potential drop across the equivalent resistance is $\Delta V_{ac}=R_{eq}I\left(t\right)=2000\times 0.00344=6.88\ {\rm V}$.  we have $\Delta V_{6k}=\Delta V_{3k}=6.88\ {\rm V}$ therefore
$I_{6k}=\frac{\Delta V_6}{R_6}=\frac{6.88}{6000}=1.148\ {\rm mA}$
$I_{3k}=\frac{\Delta V_3}{R_3}=\frac{6.88}{3000}=2.245\ {\rm mA}$

The following circuit is in its steady state; the charge on the $6\,\mu {\rm F}$ capacitor is $90\,\mu{\rm C}$. Find the
(a) value of the EMF and
(b) the charge on the $3\,\mu {\rm F}$ capacitor.

In the circuit shown, point a'' is at $-20\ {\rm V}$ with respect to ground. Find the value of the unknown EMF, the value of the unknown resistor, the magnitude and direction of the current through the $5\,\Omega$ resistor.

In the following circuit, the capacitor is initially uncharged. At $t=0$, the switch is thrown to position a''. When the voltage across the capacitor reaches $10\ {\rm V}$, the switch is automatically and simultaneously thrown to b''. Find the following quantities at $t=100\ {\rm ms}$,
(a) The charge on the capacitor
(b) The magnitude and direction of the current through each capacitor.

In the steady state circuit shown below, ($i$) the power output of the $4\,\Omega$ resistor is $324\ {\rm W}$; ($ii$) the voltage drop across the $10\ \Omega$ resistor is $50\ {\rm V}$; and ($iii$) point b'' is at a higher potential than point (a). Find:
(a) The value of the unknown EMF
(b) The magnitude of the unknown resistor
(c) The magnitude and direction of the current through it; and
(d) The charge on the capacitor. Which plate is positively charged?

What is the current through resistor $R_1$ in this circuit?

A multiloop circuit is shown on the figure. Some circuit quantities are not labeled. It is not necessary to solve the entire circuit. What is the current $I_2$?

For the circuit shown in the figure, the switch S is initially open and the capacitor voltage is $80\ {\rm V}$. The switch is then closed at time $t=0$. How long after closing the switch will the current in the resistor be $7.0\ \mu {\rm A}$?

Recall that the current in the $RC$ circuits after closing or opening the switch S is given by $I\left(t\right)=I_0{\exp \left(-\frac{t}{RC}\right)\ }$, where $I_0$ is the initial current at $t=0$. The voltage across the resistor is the same as the capacitance so $I_0=\frac{V_0}{R}=\frac{80}{2.8\times {10}^6}=2.857\times {10}^{-5}\ {\rm A}$. Therefore, by definition of the time constant $\tau$ in the $R-C$ circuits, we have
$\tau=RC=\left(2.8\times {10}^6\right)\left(22\times {10}^{-6}\right)=61.6\ {\rm s}$
$I\left(t\right)=I_0{\exp \left(-\frac{t}{RC}\right)\ }\Rightarrow \ 7\mu {\rm A}=\left(25.87\mu {\rm A}\right){\exp \left(-\frac{t}{61.6}\right)\ }$
Taking natural logarithm on both sides gives
${\ln \left(\frac{7}{25.87}\right)\ }=-\frac{t}{61.6}\Rightarrow t=86.6\ {\rm s}$

For the circuit shown in the figure, $I=0.5\ {\rm A}$ and $R=12\ \Omega$. What is the value of the emf ${\mathcal E}$?

The capacitor circuit in the figure has an $18.0\ {\rm V}$ battery connected to the capacitors with the following capacitance: $C_1=24.6\ \mu {\rm F}$ , $C_2=14.2\ \mu{\rm F}$ , $C_3=36.2\ \mu{\rm F}$.
(a) Find the equivalence capacitance of this circuit.
(b) Find the charge on each capacitor.

A capacitor of capacitance $C=4\ \mu {\rm F}$ has been charged so that the potential difference between its plates is 80\, {\rm V}. The capacitor is then connected to a $25\, {\rm k\Omega}$ resistor as shown. The switch S is closed and the capacitor begins to discharge
(a) Calculate the time constant of the circuit
(b) Determine the charge on the capacitor when the switch is closed at $t=0$
(c) Calculate the current through the resistor immediately after the switch is closed
(d) Calculate the current $10\ {\rm ms}$ after the switch is closed.
(e) Calculate the time after which the potential difference has decreased to one half its maximum value.

For the resistor circuit shown,$R_1=1\,\Omega\,R_2=R_3=R_4=6\,\Omega\,R_4=6\,\Omega$. The battery provides a potential difference of $V_0=12\,{\rm V}$.
(a) Find the equivalent resistance.
(b) Calculate the potential drop across each of the resistors and the current through each resistor.

(a) The $R_2$ and $R_3$ are in parallel, so $\frac{1}{R_{23}}=\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\Rightarrow R_{23}=2\,\Omega$.
$R_{23}$ and $R_1$ are in series so $R_{23,1}=R_{23}+R_1=2+1=3\,\Omega$
$R_{23,1}$ and $R_4$ are in parallel so $\frac{1}{R_{eq}}=\frac{1}{R_{23,1}}+\frac{1}{R_4}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\Rightarrow R_{eq}=2\,\Omega$
(b) First find the total current by $I_{tot}=\frac{V_0}{R_{eq}}=\frac{12}{2}=6\,{\rm A}$. Resistors in parallel all have the same potential drop, so
$V_{23,1}=V_4=V_0=12\ {\rm V}$
$I_{23,1}=\frac{V_{23,1}}{R_{23,1}}=\frac{12}{3}=4\ {\rm A\ ,\ \ }I_4=\frac{{V}_{4}}{{R}_{4}}=\frac{12}{6}=2\ {\rm A}$
The sum of the $I_{23,1}$ and $I_4$ must be total current in the circuit $I_{tot}$.
Resistors in series all have the same current, so
$I_{23,1}=I_{23}=I_1=4\ {\rm A}$
$V_1=I_1R_1=4\times 1=4\ {\rm V\ \ ,\ \ } V_{23}=I_{23}R_{23}=4\times 2=8\ {\rm V}$
The sum of the $V_1$ and $V_{23}$ must be the total differential potential circuit (since these are in parallel to the battery). To find the current through $R_2$ and $R_3$, since these are in parallel to each other so all have the same differential drop i.e. $V_2=V_3=8V$
$I_2=\frac{V_2}{R_2}=\frac{8}{4}=2{\rm A\ ,\ \ }I_3=\frac{V_3}{R_3}=\frac{8}{4}=2{\rm A}$
To check that we are on the right way, the sum of the $I_2$ and $I_3$ must be equal to $I_1$ because they are in series!
In summary: $I_1=4{\rm A\ ,\ } I_2=I_3=I_4={\rm 2A\ ,\ }{{\rm V}}_{{\rm 1}}{\rm =4V,\ \ }{{\rm V}}_{{\rm 2}}{\rm =}{{\rm V}}_{{\rm 3}}{\rm =8V\ ,\ }{{\rm V}}_{{\rm 4}}{\rm =12V}$

Calculate the current $I_2$ through and the voltage $\Delta V_2$ across the second resistor $R_2$ in the diagram. The other quantities are $R_1=5\,\Omega$,$R_2=4\,\Omega$ , $R_3=3\,\Omega$ , ${\mathcal E}=9\,{\rm V}$

In the circuit shown, $R=150\,\Omega$ , $R_1=R_2=10\,\Omega$ , ${{\mathcal E}}_1=12\,{\rm V}\$and ${{\mathcal E}}_2=9\,{\rm V}$. First, write down three equations that can be solved for the three currents. Then find that the current through resistor $R$.

In the circuit shown in Figure, switch S is closed at time $t=0$
(a) Find the reading of each meter just after S is closed.
(b) What does each meter read long after S is closed?

An emf source with potential difference ${\mathcal E}$ and internal resistance $r$ is connected to a resistor $R$ and a capacitor with capacitance $C$ in series connection through a switch $S$ as in the figure. Initially there is no charge in the capacitor and at $t=0$ switch $S$ is closed. Write your answers in terms of ${\mathcal E},r,R,C$.

(a) Show the direction of the positive current flow and write down the equation for charge on the capacitor for $t>0$, using Kirchhoff's rules. (note that your equation should include charge $q$, only (not current) as a dependent variable)
(b) Show that $q\left(t\right)=C{\mathcal E}\left(1-e^{-\frac{t}{\tau}}\right)$, is the solution for the equation you write in part (a) by directly substituting it into equation provided that $t$ is chosen properly. Determine $t$
(c) What is the initial current at $t=0$ when switch $S$ is closed? What is the current passing through the circuit at $t>0$?
(d) What is the maximum amount of charge that can be stored in the capacitor and when is this amount of charge accumulated in the capacitor?
(e) Plot current versus time and charge versus time graphs. Show the initial and final current/charge values in the graphs.
(f) How much energy is stored in the capacitor at some time $t_1>0$?
(g) How much heat (energy) has been generated in the resistor $R$ from $t=0$ to $t=t_1$?

(a) The switch $S$ is closed at time $t=0$. Find the time constant of this circuit. Explain your reasoning.
(b) Suppose the switch $S$ is closed for a long time and the equilibrium is reached. Then, all of a sudden, if the switch $S$ is reopened at time $t=0$, find the power dissipated through the resistor $R$ as a function of time.

(a) In the $R-L$ circuit, the quantity $\tau=L/R\$is called the time constant. This is a measure of how quickly the current reaches its final value. For example, at a time equal to $\tau$, the current has risen to about $63\%$ of its final value or in a time equal to $2\tau$, the current reaches $86\%\$of its final value.
After closing switch $S$, the current flows in the circuit. The resistors are in series so the equivalent resistor is $R_{eq}=R+r+r=R+2r$ and the circuit reduces to the following figure.
Therefore, the time constant of this circuit is
$\tau=\frac{L}{R_{eq}}=\frac{L}{R+2r}$
(b) When the switch is closed for a long time, the inductor acts like a short circuit, that is, the inductor acts like a wire with zero resistance, so the potential drop across  $R$ is zero. In this case, the current through the circuit is found, by Ohm's law, as $i=V/2r$. Immediately after the switch $S$ is opened, the current in the main branch of the circuit is zero, but in the inductor changes continuously. Applying the loop rule to the circuit below gives the current in the inductor and $R$ as time.
$-IR-L\frac{dI}{dt}=0\Rightarrow \frac{dI}{I}=-\frac{R}{L}dt$
Integrating and solving for $I$ gives
$I(t)=I_0\,{\exp \left(-\frac{R}{L}t\right)\ }$
This is the rate of change of the current after the switch $S$ is opened. $I_0$ is the current just before the switch is opened which in this case is $I_0=V/2r$. Recall that the power dissipated in the resister $R$ is $P_{diss}=RI^2(t)$. So
$P_{diss}\left(t\right)=R{\left(\frac{V}{2r}\right)}^2{\exp \left(-\frac{R}{L}\right)\ }$

Consider the $LC$ circuit shown in the figure. If the charge on the left plate of the capacitor is known to be $q=-Q_{max}/2$ and decreasing at time $t=0$, find expressions for the charge on the left plate and current as a function of time. Here, $Q_{max}$ is the maximum charge the capacitor can have in such a circuit.

Consider $LC\$circuit shown in the figure.
(a) If the current $i=I_{max}/2$ is known to flow in the clockwise direction and the charge on the left plate is known to be positive at time $t=0$
(b) If the current $i=I_{max}/2$ is known to flow in the clockwise direction and the charge on the left plate is known to be positive at time $t=T/2$, where $T$ is the period of an $LC$ circuit
Find expressions for the current and charge on the right plate of the capacitor as a function of time. Here, $I_{max}$ is the maximum current in such a circuit.

What is the current through resistor $R_3$ in this circuit?

How long does it take after the switch S is closed for the energy stored in the capacitor in the RC circuit below reach $75\%\$of its maximum value?

In the diagram $R_1>R_2>R_3$. Rank the three resistors according to the current in them, least to greatest.
(a) 1,2,3
(b) 3,2,1
(c) 1,3,2
(d) 3,1,2
(e) All are the same.

In the diagrams, all light bulbs are identical and all emf devices are identical. In which circuit (I, II, III, IV, V) will the bulbs glow with the same brightness as in circuit X?
(a) I
(b) II
(c) III
(d) IV
(e) V

The brightness of a light bulbs is determined by the current passing through it so we must find the current through each bulb in each of the above circuits.
Using Ohm's law, $I=V/R$, and series and parallel considerations, we have
In (I) the two bulbs are in series so $R_{eq}=2R$ and $I_{tot}=V/2R$. Recall that current in the series circuits is the same for all components so $i_1=i_2=V/2R$.
In(II) we have two bulbs and sources so $R_{eq}=2R\ ,\ V_{eq}=3V\Rightarrow I_{tot}=\frac{3V}{2R}$ and the current through each bulb is $i_1=i_2=3V/2R$.
In (III) $R_{eq}=3R\ ,\ V_{eq}=2V\Rightarrow I_{tot}=\frac{2}{3}\frac{V}{R}$ and $i_1=i_2=\frac{2}{3}\frac{V}{R}$.
In (IV), $R_{eq}=\frac{R}{2}\Rightarrow I_{tot}=\frac{V}{\frac{R}{2}}=2\frac{V}{R}$. But recall that in the parallel circuits, the sum of the current in the parallel branches equals to the total current through the battery so in this case $I_{tot}=i_1+i_2\Rightarrow i_1=i_2=V/R$.
In (V) the two sources are parallel connected. In this case the total voltage of the circuit is same as the one battery that is $V_{tot}=V_1=V_2$ and $R_{eq}=2R$
$\Rightarrow I_{tot}=\frac{V}{2R}\Rightarrow i_1=i_2=\frac{V}{2R}$
As you can see from straightforward calculation above, only in the (IV) circuit the current passing through each bulbs is same as the current through the bulb of the X circuit.

In the circuit shown, the capacitor is initially uncharged, and $\mathrm{V\ =\ 10\ V}$. At time $\mathrm{t\ =\ 0}$, switch S is closed. If $\tau$ denotes the time constant, the approximate current through the $3\mathrm{\Omega }$ resistor when $t=\tau /10$ is:
(a) $0.50\, \mathrm A$
(b) $0.75\, \mathrm A$
(c) $1.0\, \mathrm A$
(d) $1.5\, \mathrm A$
(e) $3.0\, \mathrm A$

Immediately after switch S in the circuit shown is closed, the current through the battery is:
(a) 0
(b) $V_0/R_1$
(c) $V_0/R_2$
(d) $V_0/(R_1+R_2)$
(e) $V_0(R_1+R_2)/R_1R_2$

Radio receivers are usually tuned by adjusting the capacitor of an $\mathrm{LC}$ circuit. If $\mathrm{C\ =\ }{\mathrm{C}}_{\mathrm{1}}$ for a frequency of $\mathrm{600\ kHz}$, then for a frequency of $\mathrm{1200\ kHz}$ one must adjust $C$ to:
(a) $C_1/2$
(b) $C_1/4$
(c) $2C_1$
(d) $4C_1$
(e) $\sqrt{2}C_1$

An RLC series circuit, connected to a source $EMF$, is at resonance. Then:
(a) the voltage across $R$ zero
(b) the voltage across $R$ equals the applied voltage
(c) the voltage across $C$ is zero
(d) the voltage across $L$ equals the applied voltage
(e) the applied voltage and current differ in phase by 90${}^\circ$

If an inductor of inductance $L$, and resistor of resistance R and a capacitor of capacitance $C$ are connected in series across the terminals of a source, an RLC series circuit forms.
Resonance in AC circuits implies a special frequency which is determined by the values of resistance, capacitance and inductance. Let us suppose an RLC series circuit is connected across an ac source of constant voltage $V$ but adjustable angular frequency $\omega$. In ac circuits there is some important concepts such as impedance which plays the role of resistance $R$ in dc circuits and is defined as the ratio of the voltage amplitude across the circuit to the current amplitude in the circuit $Z=V/I$ or
$Z=\sqrt{R^2+\left(X_L-X_C\right)^2}=\sqrt{R^2+{\left(\omega L-\left(\frac{1}{\omega C}\right)\right)}^2}$
Where $X_L=\omega L$ and $X_C=1/\omega C$ are inductive and capacitive reactances of the inductor and capacitor of the circuit, respectively. Their SI units are ohm, the same as for resistance! At resonance which is occurred at resonance angular frequency ${\omega }_0$, the inductive and capacitive reactances are equal
$X_L=X_C\Rightarrow \ {\omega }_0=\frac{1}{\sqrt{LC}}$
One of the other important properties of RLC circuits at resonance is that the total voltage across the L-C combination is exactly zero! As a result, the voltage across the resistor is equal to the source voltage.

The impedance of the circuit shown is:
(a) $21\mathrm{\Omega }$
(b) $50\ \mathrm{\Omega }$
(c) $63\ \mathrm{\Omega }$
(d) $65\ \mathrm{\Omega }$
(e) $98\ \mathrm{\Omega }$

Category : Electric Circuits

Most useful formula in Electric Circuits:

Resistors:
$\text{in series}: R_{eq}=R_1+R_2+\cdots+R_n$
$\text{in parallel}: \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n}$
Capacitor charging:
\begin{gather*}
Q\left(t\right)=C{\mathcal E}\left(1-\exp\left(-\frac{t}{\tau}\right) \right)\\
i\left(t\right)=\frac{\mathcal E}{R}\, \exp\left(-\frac{t}{RC}\right)
\end{gather*}
Capacitor discharging:
$q\left(t\right)=Q_0\,\exp\left(-\frac{t}{RC}\right)$
\begin{align*}
i=\frac{dq}{dt}&=-\frac{Q_0}{RC}\,\exp\left(-\frac{t}{RC}\right)\\
&=I_0\,\exp\left(-\frac{t}{RC}\right)
\end{align*}

Number Of Questions : 29