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RC Circuit Problems with Solution for High Schools

In this post, we want to practice some problems with RC series circuits. All problems are easy and useful for AP physics exams. 

The topic of the RC circuit involves two sections: one is related to charging a capacitor through a resistor, and the other discharging case. For a better understanding, we have separated these two parts. 


RC Circuit: Charging Capacitor 


Problem (1): An uncharged capacitor and a resistor are connected in series shown in the figure below. The battery's emf is $\mathcal E=12\,{\rm V}$, $C=8\,{\rm \mu F}$, $R=500\,{\rm k\Omega}$. After the switch is closed, find, 
(a) The time constant of the RC circuit. 
(b) The maximum charge on the capacitor.
(c) The charge on the capacitor 6 s after the switch is closed.

A capacitor and a resistor in series with a battery in an RC circuit problem

Solution: The capacitor is initially uncharged and after closing the switch, the charge is slowly collected on it through a resistor. This is an RC circuit for charging a capacitor. 

(a) The time constant $\tau$ for an RC circuit in a charging case, is defined as the time it takes the charge on a capacitor increases to about $37\%$ its final charge or is a measure of how quickly a capacitor becomes charged. \[\tau=RC\] The SI unit of time constant is seconds $s$. Thus, the time constant of this RC series circuit is determined as below \[\tau=\left(500\times 10^3\right)\left(8\times 10^{-6}\right)=4\,{\rm s}\] Hence, for this configuration, it takes about 4 s for the capacitor to reach $37\%$ its final charge.

(b) The capacitor is initially uncharged, and gradually the charge is stored on it through the resistor. In this case, at any moment the charge on the capacitor is determined by the following formula \[Q=Q_f \left(1-e^{-t/\tau}\right)\] where $Q_f=C\mathcal{E}$ is the final charge. Thus, the amount of maximum charge on this capacitor is \[Q_{max}=C\mathcal{E}=\left(8\times 10^{-6}\right)(12)=96\quad {\rm \mu C}\] 

(c) After passing about 6 seconds, the charge on the capacitor is obtained as below \begin{align*} Q&=Q_f \left(1-e^{-t/\tau}\right)\\\\&=96\left(1-e^{-6/4}\right)\\\\&=74.5\quad {\rm \mu C}\end{align*}


 

Problem (2): In the following RC circuit, the total resistance is $20\,{\rm k\Omega}$, and the battery's emf is 12 V. Suppose the time constant of this RC circuit is $18\,{\rm \mu s}$. Find,
(a) The capacitance of the circuit.
(b) The time it takes the voltage across the resistor to reach 9 V after closing the switch.

A unknown capacitor and a resistor in series with a battery

Solution: Because the source of emf is present in the circuit so the capacitor is initially uncharged and charges slowly. 

The time constant in an RC circuit is defined to be $\tau=RC$. 

(a) Given this value and solving for the unknown capacitance, we would have \[C=\frac{\tau}{R}=\frac{18\times 10^-6}{20\times 10^3}=90\times 10^{-10}\,{\rm F}\] So, the capacitance of the capacitor is $900\,{\rm nF}$. 

(b) The potential difference across a charging capacitor in an RC circuit which is proportional to the charge on it, is found by the following formula \[V=V_0 \left(1-e^{-t/\tau}\right)\] where $V_0$ is the battery's voltage or emf and $\tau$ is the time constant.

Here, the voltage at later time is given and the unknown quantity is time itself. Putting the known values into the above formula, would give \begin{gather*} 6=9\left(1-e^{-t/18}\right)\\\\\frac{6}{9}=1-e^{-t/18}\\\\\Rightarrow e^{-t/18}=1-\frac 23=\frac 13\end{gather*} Note that, we placed the time constant in terms of $\mu s$ for simplicity, and therefore the unknown time is obtained also in $\mu s$. 

Now, take natural logarithms of both sides and solve for $t$, \begin{align*} \ln\left(e^{-t/18}\right)&=\ln\left(\frac 13\right)\\\\-\frac{t}{18}&=-1.09 \\\\\Rightarrow \quad t&=19.62\quad {\rm \mu s}\end{align*} Thus, after about $20\,{\rm \mu s}$, the voltage across the capacitor drops from 12 V to 9 V. 


 

Problem (3): A 9-V battery is used to charge a $4-{\rm \mu C}$ capacitor through a resistor of $100\,{\rm \Omega}$. Find 
(a) The initial current through the circuit.
(b) The final charge on the capacitor after completely being charged.
(c) The time it takes for the capacitor to reach $96\%$ its maximum charge.

Solution: In an RC series circuit, first of all, find the time constant because all other quantities depend on it. The time constant is also the product of the resistance and the capacitor \[\tau=RC=(100)\left(4\times 10^{-6}\right)=0.4\,{\rm ms}\] 

(a) When a capacitor is being charged in an RC circuit, the current at any instant of time is found by the following formula \[I=I_0 \left(1-e^{-t/\tau}\right)\] where $I_0=\frac{\mathcal E}{R}$ is the initial current in the circuit. So, \[I_0=\frac{\mathcal E}{R}=\frac{9}{100}=0.09\,{\rm A}\] Thus, the initial current is $90\,{\rm mA}$. 

(b) In a capacitor charging case, the final charge on the capacitor is always given by $Q_{max}=C\mathcal E$. So, \[Q_{max}=\left(4\times 10^{-6}\right)(9)=36\quad{\rm \mu C}\] 
(c) We are asked the unknown time $t$ in which the charge reaches $96\%$ of its final value i.e. $Q_2=0.95Q_{max}$. The capacitor is being charged so its charge varies with time as below formula \[Q=Q_{max}\left(1-e^{-t/\tau}\right)\] Substituting the known values into the above formula, we will have \begin{align*} Q&=Q_{max}\left(1-e^{-t/\tau}\right)\\\\0.96Q_{max}&=Q_{max}\left(1-e^{-t/0.4}\right)\\\\ \Rightarrow \quad e^{-t/0.4}&=1-0.96=0.04\end{align*} Now, take natural logarithms of both sides and solve for time $t$ \begin{gather*} \ln\left(e^{-t/0.4}\right)=\ln(0.04)\\\\-\frac{t}{0.4}=-3.21 \\\\\Rightarrow \quad t=1.284\,{\rm ms}\end{gather*} Hence, it takes about 1.3 ms for the capacitor  charges to $96\%$ of its final value. 


 

Problem (4): An initially uncharged $8-{\rm \mu F}$ capacitor is placed in series with a resistance of $8\,{\rm \Omega}$ and a source of $12\,{\rm V}$. Find, 
(a) The initial and final voltage across the resistor.
(b) The initial and final voltage across the capacitor.

Solution: suppose an uncharged capacitor in an RC circuit along with a switch. Just after closing the switch, the capacitor always behaves like a typical wire. Consequently, at this time $t=0$, there are only a resistor and a battery. So, the voltage across the resistor is equal to that of the battery. Hence, at time $t=0$, we have \[\Delta V_R=V_{battery}=12\,{\rm V}\]

Gradually the charge is stored on the capacitor and makes a voltage drop across it. After a long time, when the capacitor is fully charged, the current through the resistor becomes zero. So, using Ohm's law, $\Delta V_R=IR$, the voltage difference across the resistor is also zero. 

(b) As mentioned above, at time $t=0$, immediately after closing the switch, the uncharged capacitor is like a wire. So, the voltage difference across it, at this moment, is zero. 

After a long time, the current becomes zero and thus the resistor eliminates from the circuit. So, the voltage drop across the capacitor is the same voltage as the source. \[\Delta V_C= V_{battery}=12\,{\rm V}\]



Problem (5): In an RC circuit, there are two $4\,{\rm \mu F}$-uncharged capacitors, two $2.5\,{\rm k\Omega}$-resistors, and a source of 24 V. All these components are connected in series. After closing the switch,
(a) Find the initial current through the resistors?
(b) How long does it take the current to drop from its initial value to $2.4\,{\rm mA}$?

Two capacitor and two resistor in series with a battery in an RC circuit problem

Solution: In an RC series circuit, we have one capacitor, one resistor, and one source. First, we must find the equivalent capacitance and resistance of the above circuit to convert it into one capacitor and one resistor. 

The equivalent resistance of two resistors in series is the sum of their resistances, so \[R=R_1+R_2=5\quad {\rm k\Omega}\] The equivalent capacitance of two capacitors in series is found as formula below \[\frac 1C=\frac 1C_1+\frac 1C_2=2\left(\frac 14\right)=\frac 12\] Inverting the above, gives the equivalent capacitance of two capacitor in series. So, $C=2\,{\rm \mu F}$. 

(a) Just after closing the switch $S$, the capacitor doesn't have any charge, so it is behavior like a typical wire. In fact, at this time, there are only a resistor and a source, which using Ohm's law formula, we can find the current through the resistor as \[I=\frac{V_0}{R}=\frac{24}{2.5\times 10^3}=0.0096\quad {\rm A}\] Thus, the current through the resistors initially is $9.6\,{\rm mA}$. 

(b) In an RC series circuit, when the capacitor is being charged, the current reduces with time as below formula \[I=I_0e^{-t/RC}\] where $I_0$ is the initial current just after closing the switch. 

The current at the unknown time $t$ is given as 2.4 mA, substituting this into the above formula, we have \[2.4=9.6e^{-t/RC}\] where we used the initial current $I_0=9.6\,{\rm mA}$ computed in part (a). 

Now, take natural logarithm of both sides and solve for the time $t$, gives \begin{gather*} \ln\left(\frac{2.4}{9.6}\right)=\ln e^{-t/RC}=-\frac{t}{RC}\\\\ \Rightarrow \quad t=1.38RC\end{gather*} where $\tau=RC$ is the time constant of a RC circuit. The time constant of this circuit is \[\tau=(5000)\times \left(2\times 10^{-6}\right)=0.01\,{\rm s}\] Thus, the time it takes for the current to drop from initially 9.6 mA to 2.4 mA is \[t=1.38\tau=1.38\times 0.01=13.8\,{\rm ms}\] 


RC Circuit: Discharging Capacitor

 

Problem (6): A $5\,{\rm \mu F}$-capacitor is charged to a 12 V and then connected to a $400\,{\rm \Omega}$-resistor. Find, 
(a) The time constant of this RC circuit.
(b) The initial charge stored on the capacitor.
(c) The initial current through the resistor.
(d) The charge on the capacitor after 3 ms.

Dischargin a capacitor through a resistor in an RC circuit problem

Solution: A fully charged capacitor is connected to a resistor and consequently discharges through it. In this case, there is no battery in the circuit.

(a) The time constant, $\tau=RC$, is the time it takes for the charges on the capacitor to decrease to about $37\%$ of its initial charges. For this RC circuit, we have \[\tau=400 \times \left(5\times 10^{-6}\right)=2\times 10^{-3}\,{\rm s}\] Thus, the time constant of this RC circuit is $2\,{\rm ms}$. 

(b) Just after closing the switch, the initial charges on the capacitor is \[Q_0=CV_0=\left(5\times 10^-6 \right)(12)=60\,{\rm \mu C}\]
(c) When a capacitor in an RC series circuit is being discharged, the current at any moment is determined by the following formula \[I=I_0 e^{-t/\tau}\] where $I_0=\frac{V_0}{R}$ is the initial current at time $t=0$. Thus, \[I_0=\frac{V_0}{R}=\frac{12}{400}=0.03\quad {\rm A}\] Multiplying by $1000$ gives the current in terms of milliamperes. So, $I_0=30\,{\rm mA}$. 

(d) In the discharging case, the charge on the capacitor varies with time as $Q=Q_0e^{-t/RC}$. So, after $t=3\,{\rm ms}$, the amount of remaining charge is \begin{align*} Q&=Q_0e^{-t/RC}\\\\&=\left(60\times 10^{-6}\right)e^{-\frac{3}{2}}\\\\&=13.38\,{\rm \mu C}\end{align*} 



Problem (7): A $C=4\,{\rm \mu F}$-capacitor is charged to a source of $\mathcal{E}=24\,{\rm V}$ and then connected to a resistor of $R=200\,{\rm \Omega}$ in series. After the switch is closed, find
(a) The initial current through the resistor.
(b) The current after passing two time constants.
(c) The time constant of the circuit.

Solution: After closing the switch, the capacitor discharges (loses its charges) through the resistor. Recall that the time rate of the electric charge is defined as current. 

The exact form of this variation with time is $I=I_0e^{-t/RC}$, where $I_0=\frac{\mathcal{E}}{R}$ is the initial current in the RC circuit. 

(a) Immediately after closing the switch, the current in the RC circuit is found by the following formula \[I_0=\frac{\mathcal{E}}{R}=\frac{24}{200}=0.12\,{\rm A}\]
(b) The capacitor is being discharged, so the current at any instant of time in the RC circuit is determined as $I=I_0e^{-t/\tau}$ where $\tau=RC$ is the time constant of the circuit. 

Thus, the current after 2 time constants, $t=2\tau$, is \begin{align*} I&=I_0e^{-t/RC}\\\\&=0.12\,e^{-2\tau/\tau}=0.0162\quad {\rm A}\end{align*} Hence, after passing 2 time constants, the current is about 16.2 mA. 

The time constant is calculated as below \[\tau=200\times \left(4\times 10^-6 \right)=8\times 10^{-4}\,{\rm s}\] This is the time duration in which the charges on the capacitor decrease to about $37\%$ of its initial charges. 


 

Problem (8): Suppose an RC series circuit with a capacitor of $5\,{\rm \mu F}$, a resistor of $80\,{\rm k\Omega}$, and an initial voltage difference across the capacitor of $6\,{\rm V}$. 
(a) Find the time constant of this RC circuit.
(b) How long does it take the capacitor to lose half its initial charge?
(c) How long does it take the capacitor to lose $99.99\%$ of its initial charge?

Solution: The capacitor has an initial voltage across itself so the capacitor is fully charged initially and discharges through the resistor slowly.

(a) The time constant $\tau$ for a discharging capacitor in an RC circuit tells us how much time is required the charge on the capacitor reaches from its maximum value to about $37\%$ of that value. 

In an RC series circuit the product of the resistance $R$ times the capacitance $C$ is defined as time constant. So, \[\tau=RC=\left(80\times 10^3\right)\left(5\times 10^{-6}\right)=0.4\,{\rm s}\] 
(b) ``how long'' means the time is unknown. Let initial charge be $Q_0$, then we want some later time the charge to be $Q=(0.5)Q_0$.

Recall that for a discharging capacitor the charge at any instant of time is given by the following formula \[Q=Q_0 e^{-t/\tau}\] Putting the known values into it, we will have \begin{align*} 0.5Q_0 &=Q_0 e^{-t/0.4}\\\\ \ln(0.5)&=\ln\left(e^{-t/0.4}\right)=-\frac{t}{0.4}\\\\ \Rightarrow \quad t&=(0.4)\ln(0.5)\\\\&=0.277\quad{\rm s}\end{align*} Hence, it takes about 0.3 s the charge on the capacitor becomes half its initial value.

(c) ``loses $99.99\%$ of its initial value'' means that the remaining charge on the capacitor is only $1-0.9999$ of its initial value. It means that $Q=0.0001Q_0$. So, \begin{align*} 0.9999Q_0 &=Q_0 e^{-t/0.4}\\\\ \ln(0.0001)&=\ln\left(e^{-t/0.4}\right)=-\frac{t}{0.4}\\\\ \Rightarrow \quad t&=(0.4)\ln(0.0001)\\\\&=3.684\quad{\rm s}\end{align*} This calculation gives us an estimate of how much it takes the capacitor to be completely discharged. 


 

Problem (9): A $150\,{\rm \mu F}$-capacitor is charged to a 1500 V and then connected in series with an unknown resistor. It is observed that the capacitor loses $95\%$  of its charge in 50 ms. Find the resistance of the resistor.

Solution: The capacitor is fully charged initially, and then discharges through the resistor. In this case, the initial charge $Q_0$ on the capacitor decreases with time as below formula \[Q=Q_0 e^{-t/RC}\] where $Q$ is the charge on the capacitor at any instant of time. 

It is said in the question that after 50 ms the capacitor loses $95\%$ of its initial charge. This means that after this time duration, the charge on the capacitor is only 5 of its initial charge, $Q=0.05Q_0$. So, we have \begin{align*} Q&=Q_0 e^{-t/RC}\\\\ 0.05Q_0&=Q_0 e^{-t/RC}\\\\0.05&=e^{-t/RC}\end{align*} Taking natural logarithm and solving for the unknown resistance $R$, would give \begin{align*} \ln (0.05)&=\ln\left(e^{-t/RC}\right)=-\frac{t}{RC}\\\\ \Rightarrow \quad R&=-\frac{t}{C\ln(0.05)}\\\\ &=-\frac {50\times 10^{-3}}{\left(150\times 10^{-6}\right)\times \ln(0.05)}\\\\ &=111.26 \quad {\rm \Omega} \end{align*}


 

Problem (10): In an RC series circuit, a capacitor C is being discharged through a resistor of R. How long does it take for the charge on the capacitor to drop to one-third its initial value? 

Solution: ``how long'' means time duration. In an RC circuit capacitor discharging, the charge on the capacitor at any time instant is given by formula \[Q=Q_0 e^{-t/RC}\] where $Q_0$ represents the maximum charge. We want to find when $Q=\frac 13 Q_0$. Substituting this relation into above formula, we have. \begin{gather*} Q=Q_0e^{-t/RC}\\\\\frac 13 Q_0=Q_0e^{-t/RC}\\\\ \frac 13 =e^{-t/RC}\end{gather*} Now, take natural logarithms of both sides and then solve for the unknown time $t$ \begin{gather*} \ln\left(\frac 13\right)=\ln e^{-t/RC}=-\frac{t}{RC}\\\\ \Rightarrow t=-RC\ln\left(\frac 13\right)=1.79RC\end{gather*} Thus, after passing about 1.8 of time constant $\tau$ the charge on the capacitor reaches one-sixth its initial value. (Remind that the time constant is defined as $\tau=RC$)

Other related topics: 

RL circuit problems with solutions
Electric circuits problems and answers



Author: Ali Nemati
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