Calculus

Integration of Functions

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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.

    A series RC circuit
    Figure 1.

    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 current change in an RC circuit.
    Figure 2.

    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.

    A series RL circuit
    Figure 3.

    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:

    \[I\left( t \right) = \frac{\varepsilon }{R}\left( {1 – {e^{ – \frac{R}{L}t}}} \right).\]

    The current change in an RL circuit.
    Figure 4.

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

    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:

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

    An electric circuit with capacitors.
    Figure 5.

    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:

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

    An electric circuit with inductors.
    Figure 6.

  • Solved Problems

    Click 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 5

    The current and voltage across a circuit element are given by the equations \[{I\left( t \right) = 20{e^{ – 100t}}\,\left( {mA} \right),\;\;}\kern0pt{V\left( t \right) = 10 – 5{e^{ – 100t}}\,\left( V \right).}\] Determine the total energy dissipated by the element between \(t = 0\) and \(t = 10\,{ms}.\)

    Example 6

    At time \(t = 0,\) the emf \(\varepsilon = 50\,V\) is applied to the initially uncharged capacitor \(C = 10\,\mu F.\) The capacitor starts to charge through the resistor \(R = 100\,k\Omega.\) Determine the number of electrons on the negative capacitor’s plate in \(1\) second.

    Example 7

    The current and voltage across a circuit element change according to a sinusoidal law: \[{I\left( t \right) = {I_0}\sin \left( {\frac{{2\pi t}}{T} + \theta } \right),\;\;}\kern0pt{V\left( t \right) = {V_0}\sin \left( {\frac{{2\pi t}}{T}} \right),}\] where \(T\) is the period of oscillations, \(\theta\) is the phase difference, \({I_0}\) and \({V_0}\) are initial values of the current and voltage. Find the average power dissipated in the circuit element over one cycle period.

    Example 8

    A source of constant emf \(\varepsilon = 100\,V\) is connected to a circuit with an initial resistance of \({R_0} = 20\,\Omega.\) Calculate the charge \(Q\) that will pass in the circuit during \(T = 1\,min,\) if the resistance increases linearly at a rate of \(\alpha = 1 \large{\frac{\Omega}{s}}\normalsize.\)

    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.

    Charging a capacitor in an RC circuit.
    Figure 7.

    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.

    A source-free RL circuit.
    Figure 8.

    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:

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

    Example 5.

    The current and voltage across a circuit element are given by the equations \[{I\left( t \right) = 20{e^{ – 100t}}\,\left( {mA} \right),\;\;}\kern0pt{V\left( t \right) = 10 – 5{e^{ – 100t}}\,\left( V \right).}\] Determine the total energy dissipated by the element between \(t = 0\) and \(t = 10\,{ms}.\)

    Solution.

    The instantaneous power dissipated by the element is

    \[P\left( t \right) = I\left( t \right)V\left( t \right).\]

    To find the total energy \(E,\) we integrate \(P\left( t \right)\) over the time interval from \(t = 0\) to \(t = 10\,{ms} = 0.01 \,{sec}:\)

    \[{E = \int\limits_0^{0.01} {P\left( t \right)dt} }={ \int\limits_0^{0.01} {I\left( t \right)V\left( t \right)dt} }={ \int\limits_0^{0.01} {20{e^{ – 100t}}\left( {10 – 5{e^{ – 100t}}} \right)dt} }={ \int\limits_0^{0.01} {\left( {200{e^{ – 100t}} – 100{e^{ – 200t}}} \right)dt} }={ \left. {\left[ {\frac{{200{e^{ – 100t}}}}{{\left( { – 100} \right)}} – \frac{{100{e^{ – 200t}}}}{{\left( { – 200} \right)}}} \right]} \right|_0^{0.01} }={ \left. {\left( {\frac{1}{2}{e^{ – 200t}} – 2{e^{ – 100t}}} \right)} \right|_0^{0.01} }={ \left( {\frac{1}{{2{e^2}}} – \frac{2}{e}} \right) – \left( {\frac{1}{2} – 2} \right) }={ \frac{{1 – 4e + 3{e^2}}}{{2{e^2}}} }\approx{ 0.83\,{mJ}}\]

    The answer is represented in \(mJ\) because the current is measured in \(mA\) and voltage is measured in \(V.\)

    Example 6.

    At time \(t = 0,\) the emf \(\varepsilon = 50\,V\) is applied to the initially uncharged capacitor \(C = 10\,\mu F.\) The capacitor starts to charge through the resistor \(R = 100\,k\Omega.\) Determine the number of electrons on the negative capacitor’s plate in \(1\) second.

    Solution.

    When the capacitor is charging, the current in the circuit changes according to the exponential law:

    \[I\left( t \right) = {I_0}{e^{ – \frac{t}{\tau }}} = \frac{\varepsilon }{R}{e^{ – \frac{t}{{RC}}}}.\]

    The charge on the capacitor in \(1\,s\) is

    \[{Q = \int\limits_0^1 {I\left( t \right)dt} }={ \frac{\varepsilon }{R}\int\limits_0^1 {{e^{ – \frac{t}{{RC}}}}dt} }={ \frac{\varepsilon }{R}\left. {\left[ {\left( { – RC} \right){e^{ – \frac{t}{{RC}}}}} \right]} \right|_0^1 }={ \varepsilon C\left( {1 – {e^{ – \frac{1}{{RC}}}}} \right) }={ 50 \times {10^{ – 5}} \times \left( {1 – {e^{ – \frac{1}{{{{10}^5} \times {{10}^{ – 5}}}}}}} \right) }={ 3.16 \times {10^{ – 4}}\,\left({C}\right)}\]

    Given that the charge of an electron is

    \[e = 1.6 \times {10^{ – 19}}\,C,\]

    we find the amount of electrons on the capacitor’s plate:

    \[{N = \frac{Q}{e} = \frac{{3.16 \times {{10}^{ – 4}}}}{{1.6 \times {{10}^{ – 19}}}} }={ 1.97 \times {10^{15}}}\]

    Example 7.

    The current and voltage across a circuit element change according to a sinusoidal law: \[{I\left( t \right) = {I_0}\sin \left( {\frac{{2\pi t}}{T} + \theta } \right),\;\;}\kern0pt{V\left( t \right) = {V_0}\sin \left( {\frac{{2\pi t}}{T}} \right),}\] where \(T\) is the period of oscillations, \(\theta\) is the phase difference, \({I_0}\) and \({V_0}\) are initial values of the current and voltage. Find the average power dissipated in the circuit element over one cycle period.

    Solution.

    The average power over one period \(T\) is given by the integral

    \[{\bar P = \frac{1}{T}\int\limits_0^T {P\left( t \right)dt} }={ \frac{1}{T}\int\limits_0^T {I\left( t \right)V\left( t \right)dt} .}\]

    Substituting the expressions for the current \(I\left( t \right)\) and voltage \(V\left( t \right),\) we obtain:

    \[{\bar P \text{ = }}\kern0pt{\frac{{{I_0}{V_0}}}{T}\int\limits_0^T {\sin \left( {\frac{{2\pi t}}{T} + \theta } \right)\sin \left( {\frac{{2\pi t}}{T}} \right)dt} .}\]

    Using the product-to-sum identity

    \[{\sin \alpha \sin \beta \text{ = }}\kern0pt{\frac{1}{2}\left[ {\cos \left( {\alpha – \beta } \right) – \cos \left( {\alpha + \beta } \right)} \right],}\]

    we can rewrite the integrand in the form

    \[{\sin \left( {\frac{{2\pi t}}{T} + \theta } \right)\sin \left( {\frac{{2\pi t}}{T}} \right) }={ \frac{1}{2}\left[ {\cos \left( { – \theta } \right) – \cos \left( {\frac{{4\pi t}}{T} + \theta } \right)} \right] }={ \frac{1}{2}\left[ {\cos \theta – \cos \left( {\frac{{4\pi t}}{T} + \theta } \right)} \right].}\]

    Hence,

    \[{\bar P \text{ = }}\kern0pt{\frac{{{I_0}{V_0}}}{{2T}}\int\limits_0^T {\left[ {\cos \theta – \cos \left( {\frac{{4\pi t}}{T} + \theta } \right)} \right]dt} }={ \frac{{{I_0}{V_0}}}{{2T}}\left. {\left[ {t\cos \theta – \frac{{T\sin \left( {\frac{{4\pi t}}{T} + \theta } \right)}}{{4\pi }}} \right]} \right|_0^T }={ \frac{{{I_0}{V_0}}}{2}\left[ {\cos \theta – \underbrace {\frac{{\sin \left( {4\pi + \theta } \right) – \sin \theta }}{{4\pi }}}_0} \right] }={ \frac{{{I_0}{V_0}\cos \theta }}{2}.}\]

    As you can see, the maximum average power is achieved at \(\theta = 0:\)

    \[{{\bar P}_{\max }} = \frac{{{I_0}{V_0}}}{2}.\]

    Example 8.

    A source of constant emf \(\varepsilon = 100\,V\) is connected to a circuit with an initial resistance of \({R_0} = 20\,\Omega.\) Calculate the charge \(Q\) that will pass in the circuit during \(T = 1\,min,\) if the resistance increases linearly at a rate of \(\alpha = 1 \large{\frac{\Omega}{s}}\normalsize.\)

    Solution.

    The resistance \(R\) of the circuit changes according to the law

    \[R\left( t \right) = {R_0} + \alpha t.\]

    By the Ohm’s law,

    \[I\left( t \right) = \frac{\varepsilon }{{R\left( t \right)}} = \frac{\varepsilon }{{{R_0} + \alpha t}}.\]

    The find the charge \(Q,\) we integrate the current \(I\left( t \right)\) over the time interval \(\left[ {0,T} \right],\) where \(T = 1\,min = 60\,s.\) This yields:

    \[{Q = \int\limits_0^T {I\left( t \right)dt} }={ \int\limits_0^T {\frac{{\varepsilon dt}}{{{R_0} + \alpha t}}} }={ \varepsilon \int\limits_0^T {\frac{{dt}}{{{R_0} + \alpha t}}} }={ \frac{\varepsilon }{\alpha }\left. {\ln \left( {{R_0} + \alpha t} \right)} \right|_0^T }={ \frac{\varepsilon }{\alpha }\left[ {\ln \left( {{R_0} + \alpha T} \right) – \ln {R_0}} \right] }={ \frac{\varepsilon }{\alpha }\ln \frac{{{R_0} + \alpha T}}{{{R_0}}} }={ \frac{\varepsilon }{\alpha }\ln \left( {1 + \frac{{\alpha T}}{{{R_0}}}} \right).}\]

    By substituting the specified values, we get

    \[{Q = \frac{\varepsilon }{\alpha }\ln \left( {1 + \frac{{\alpha T}}{{{R_0}}}} \right) }={ \frac{{100}}{1}\ln \left( {1 + \frac{{1 \times 60}}{{20}}} \right) }\approx{ 138.6\,C}\]