# Integrals in Electric Circuits

Derivatives and integrals are widely used to describe transient processes in electric circuits. Below, we look at some typical problems that can be solved using integration. We confine ourselves to consideration of first order circuits.

## Relationship Between Charge and Current

Electric current $$I$$ is defined as the rate of flow of charge $$Q$$ and is expressed by the derivative

$I\left( t \right) = \frac{{dQ\left( t \right)}}{{dt}}.$

Turning the equation the other way round, we get the integral formula

$Q = \int\limits_{{t_1}}^{{t_2}} {I\left( t \right)dt} ,$

which represents the amount of charge passing through the wire between the times $$t = {t_1}$$ and $$t = {t_2}.$$

## RC Circuit

A simple series RC Circuit is an electric circuit composed of a resistor and a capacitor.

After the switch is closed at time $$t = 0,$$ the current begins to flow across the circuit. The voltage across the resistor is given by the Ohm's law:

${V_R}\left( t \right) = I\left( t \right)R.$

The voltage across the capacitor is expressed by the integral

${V_C}\left( t \right) = \frac{1}{C}\int\limits_0^t {I\left( s \right)ds} ,$

where $$C$$ is a capacitance value, $$s$$ is the internal variable of integration.

By the Kirchhoff's voltage law (KVL), we can write

${V_R}\left( t \right) + {V_C}\left( t \right) = \varepsilon ,$

where $$\varepsilon$$ is the electromotive force (emf) of the power supply (we assume that $$\varepsilon$$ is constant).

Hence,

$RI\left( t \right) + \frac{1}{C}\int\limits_0^t {I\left( s \right)ds} = \varepsilon .$

By differentiating with respect to $$t,$$ we can convert this integral equation into a linear differential equation:

$R\frac{{dI}}{{dt}} + \frac{1}{C}I\left( t \right) = 0,$

which has the solution in the form

$I\left( t \right) = \frac{\varepsilon }{R}{e^{ - \frac{t}{{RC}}}}.$

The time constant $$\tau = RC$$ here determines how quickly the transient process in the circuit occurs.

## RL Circuit

A simple RL Circuit has a resistor and an inductor connected in series.

When the switch at time $$t = 0$$ is closed, a constant emf $$\varepsilon$$ is applied and the current $$I$$ begins to flow across the circuit.

Similarly to the previous section, the voltage across the resistor is given by

${V_R}\left( t \right) = I\left( t \right)R.$

The voltage across the inductor is expressed by the derivative

${V_L}\left( t \right) = L\frac{{dI}}{{dt}}.$

So, by KVL,

${V_R}\left( t \right) + {V_L}\left( t \right) = \varepsilon ,$

or

$RI\left( t \right) + L\frac{{dI}}{{dt}} = \varepsilon .$

Integrating this linear differential equation with the initial condition $$I\left( {t = 0} \right) = 0$$ gives the following solution:

$I\left( t \right) = \frac{\varepsilon }{R}\left( {1 - {e^{ - \frac{R}{L}t}}} \right).$

We see that the time constant for an RL circuit is given by $$\tau = \frac{L}{R}.$$

## Power and Energy

Electric energy $$E,$$ measured in joules (J), is a form of energy that results from kinetic or potential energy possessed by electric charges.

Electric power $$P,$$ measured in watts (W), is the rate at which electric energy is transferred by an electric circuit.

The power dissipated in a direct current $$\left({DC}\right)$$ circuit element is given by the formula

$P = VI,$

where $$V$$ is the voltage across the element and $$I$$ is the current in the circuit.

In particular, if the power is dissipated in a resistor of resistance $$R,$$ then

$P = VI = {I^2}R = \frac{{{V^2}}}{R}.$

The energy dissipated by a $$DC$$ circuit element during the time period $$\left[ {0,t} \right]$$ is given by

$E = VIt.$

When the voltage and current change in time, the instantaneous power is defined as

$P\left( t \right) = V\left( t \right)I\left( t \right).$

In this case, the energy dissipated during the time interval $$\left[ {0,t} \right]$$ is given by the integral

$E = \int\limits_0^t {V\left( s \right)I\left( s \right)ds} ,$

where $$s$$ is the internal variable of integration.

## Energy Stored in a Capacitor

Moving a small charge $$dq$$ from one plate of a capacitor to the other requires the work

$dW = Vdq = \frac{q}{C}dq,$

where $$C$$ is the capacitance and $$q$$ is the current charge of the capacitor.

Integrating from $$q = 0$$ to $$q = Q$$ gives the total energy stored in the capacitor:

${E_C} = \int\limits_0^Q {dW} = \int\limits_0^Q {\frac{q}{C}dq} = \frac{{{Q^2}}}{{2C}} = \frac{{C{V^2}}}{2}.$

## Energy Stored in an Inductor

Increasing the current in an inductor by a small value of $$di$$ requires the work

$dW = Pdt = - \varepsilon idt = iL\frac{{di}}{{dt}}dt = Lidi.$

Integrating from $$i = 0$$ to $$i = I$$ gives the total energy stored in the inductor:

${E_L} = \int\limits_0^I {dW} = \int\limits_0^I {Lidi} = \frac{{L{I^2}}}{2}.$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

The current in a circuit element varies according to the law $I(t) = \begin{cases} 2t, & 0 \lt t \le 3 \\ {t^2} - 4, & t \gt 3 \end{cases},$ where the current $$I$$ is measured in $$A$$, and time $$t$$ is measured in $${sec}.$$ Find the total charge that has entered the element by time $$T = 6\,s.$$

### Example 2

The current in a circuit increases linearly in time as $$I\left( t \right) = \alpha t$$ during the time interval $$\left[ {0,T} \right]$$ and causes the resistor $$R$$ to heat up. Assuming that the heating process is adiabatic, determine how the change in temperature of the resistor $$\Delta T$$ depends on the rate $$\alpha.$$ The specific heat capacity of the resistor material is $$c,$$ the mass of the resistor is $$m.$$

### Example 3

Suppose a capacitor $$C$$ is being charged using a source with a constant emf $$\varepsilon.$$ Calculate the thermal energy dissipated by the resistor $$R$$ over the charging time.

### Example 4

When the switch is closed at time $$t = 0,$$ the initial current in a source-free $$RL$$ circuit is $${I_0} = 1\,A.$$ Find the energy $${E_R}$$ dissipated by the resistor between $$t = 0$$ and $$T = 1\,ms,$$ if $$R = 50\,k\Omega,$$ $$L = 0.1\,H.$$

### Example 1.

The current in a circuit element varies according to the law $I(t) = \begin{cases} 2t, & 0 \lt t \le 3 \\ {t^2} - 4, & t \gt 3 \end{cases},$ where the current $$I$$ is measured in $$A$$, and time $$t$$ is measured in $${sec}.$$ Find the total charge that has entered the element by time $$T = 6\,s.$$

Solution.

The total charge $$Q,$$ measured in coulombs $$\left( C \right)$$, is obtained by integration of the current $$I\left( t \right)$$ over the time interval $$\left[ {0,T} \right]:$$

$Q = \int\limits_0^T {I\left( t \right)dt} = \int\limits_0^3 {2tdt} + \int\limits_3^6 {\left( {{t^2} - 4} \right)dt} = \left. {{t^2}} \right|_0^3 + \left. {\left( {\frac{{{t^3}}}{3} - 4t} \right)} \right|_3^6 = 9 + \left( {\frac{{216}}{3} - 4} \right) - \left( {3 - 12} \right) = 60\,C.$

### Example 2.

The current in a circuit increases linearly in time as $$I\left( t \right) = \alpha t$$ during the time interval $$\left[ {0,T} \right]$$ and causes the resistor $$R$$ to heat up. Assuming that the heating process is adiabatic, determine how the change in temperature of the resistor $$\Delta T$$ depends on the rate $$\alpha.$$ The specific heat capacity of the resistor material is $$c,$$ the mass of the resistor is $$m.$$

Solution.

The power delivered to the resistor is given by

$P\left( t \right) = {I^2}\left( t \right)R = {\alpha ^2}{t^2}R.$

The thermal energy dissipated in the element during the time interval $$\left[ {0,T} \right]$$ can be found through integration:

$E = \int\limits_0^T {P\left( t \right)dt} = \int\limits_0^T {{\alpha ^2}{t^2}Rdt} = {\alpha ^2}R\int\limits_0^T {{t^2}dt} = \frac{{{\alpha ^2}R{T^3}}}{3}.$

Since the process is adiabatic, we can write the following energy balance equation:

$\frac{{{\alpha ^2}R{T^3}}}{3} = cm\Delta \theta .$

Hence

$\Delta \theta = \frac{{{\alpha ^2}R{T^3}}}{{3cm}} \sim {\alpha ^2}.$

Thus, the change in temperature $$\Delta \theta$$ is proportional to the current rate $$\alpha$$ squared.

### Example 3.

Suppose a capacitor $$C$$ is being charged using a source with a constant emf $$\varepsilon.$$ Calculate the thermal energy dissipated by the resistor $$R$$ over the charging time.

Solution.

As the capacitor is charging, the current in the circuit varies according to the law

$I\left( t \right) = \frac{\varepsilon }{R}{e^{ - \frac{t}{{RC}}}}.$

The power loss in the resistor $$R$$ is given by

$P\left( t \right) = {I^2}\left( t \right)R = {\left( {\frac{\varepsilon }{R}{e^{ - \frac{t}{{RC}}}}} \right)^2}R = \frac{{{\varepsilon ^2}}}{R}{e^{ - \frac{{2t}}{{RC}}}}.$

The total energy $${E_R}$$ lost on the resistor over the charging period can be found through integration:

${E_R} = \int\limits_0^\infty {P\left( t \right)dt} = \frac{{{\varepsilon ^2}}}{R}\int\limits_0^\infty {{e^{ - \frac{{2t}}{{RC}}}}dt} = \frac{{{\varepsilon ^2}}}{R}\left. {\left[ {\left( { - \frac{{RC}}{2}} \right){e^{ - \frac{{2t}}{{RC}}}}} \right]} \right|_0^\infty = \frac{{{\varepsilon ^2}C}}{2}.$

Note that the energy $${E_R}$$ dissipated by the resistor while charging the capacitor is equal to the energy $${E_C}$$ stored in the capacitor:

${E_R} = {E_C} = \frac{{{\varepsilon ^2}C}}{2}.$

### Example 4.

When the switch is closed at time $$t = 0,$$ the initial current in a source-free $$RL$$ circuit is $${I_0} = 1\,A.$$ Find the energy $${E_R}$$ dissipated by the resistor between $$t = 0$$ and $$T = 1\,ms,$$ if $$R = 50\,k\Omega,$$ $$L = 0.1\,H.$$

Solution.

By the KVL, we can write:

$L\frac{{dI}}{{dt}} + IR = 0.$

This differential equation has the solution

$I\left( t \right) = {I_0}{e^{ - \frac{{Rt}}{L}}}.$

The instantaneous power delivered to the resistor is given by

$P\left( t \right) = {I^2}\left( t \right)R = I_0^2 R\,{e^{ - \frac{{2Rt}}{L}}}.$

By integrating the power from $$t = 0$$ to $$T = 1\,ms = 0.001\,s,$$ we find the energy dissipated in the resistor during the time interval:

${E_R} = \int\limits_0^T {P\left( t \right)dt} = I_0^2R\int\limits_0^T {{e^{ - \frac{{2Rt}}{L}}}dt} = \frac{{I_0^2L}}{2}\left( {1 - {e^{ - \frac{{2RT}}{L}}}} \right).$

Substituting the known values, we get:

${E_R} = \frac{{{1^2} \times 0.1}}{2}\left( {1 - {e^{ - \frac{{2 \times 50 \times 0.001}}{{0.1}}}}} \right) = \frac{1}{{20}}\left( {1 - \frac{1}{e}} \right) = \frac{{e - 1}}{{20e}} \approx 0.0316\,J = 31.6\,mJ$

See more problems on Page 2.