Integrals in Electric Circuits
Solved Problems
Example 5.
The current and voltage across a circuit element are given by the equations
\[I\left( t \right) = 20{e^{ - 100t}}\,\left( {mA} \right),\;V\left( t \right) = 10 - 5{e^{ - 100t}}\,\left( V \right).\]
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:
\[I\left( t \right) = {I_0}\sin \left( {\frac{{2\pi t}}{T} + \theta } \right),\;V\left( t \right) = {V_0}\sin \left( {\frac{{2\pi t}}{T}} \right),\]
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:
\[\bar P = \frac{{{I_0}{V_0}}}{T}\int\limits_0^T {\sin \left( {\frac{{2\pi t}}{T} + \theta } \right)\sin \left( {\frac{{2\pi t}}{T}} \right)dt} .\]
Using the product-to-sum identity
\[\sin \alpha \sin \beta = \frac{1}{2}\left[ {\cos \left( {\alpha - \beta } \right) - \cos \left( {\alpha + \beta } \right)} \right],\]
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,
\[\bar P = \frac{{{I_0}{V_0}}}{{2T}}\int\limits_0^T {\left[ {\cos \theta - \cos \left( {\frac{{4\pi t}}{T} + \theta } \right)} \right]dt} = \frac{{{I_0}{V_0}}}{{2T}}\left. {\left[ {t\cos \theta - \frac{{T\sin \left( {\frac{{4\pi t}}{T} + \theta } \right)}}{{4\pi }}} \right]} \right|_0^T = \frac{{{I_0}{V_0}}}{2}\left[ {\cos \theta - \underbrace {\frac{{\sin \left( {4\pi + \theta } \right) - \sin \theta }}{{4\pi }}}_0} \right] = \frac{{{I_0}{V_0}\cos \theta }}{2}.\]
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 \(\alpha = 1 \frac{\Omega}{s}.\)
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,
\[I\left( t \right) = \frac{\varepsilon }{{R\left( t \right)}} = \frac{\varepsilon }{{{R_0} + \alpha t}}.\]
To 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
\[Q = \frac{\varepsilon }{\alpha }\ln \left( {1 + \frac{{\alpha T}}{{{R_0}}}} \right) = \frac{{100}}{1}\ln \left( {1 + \frac{{1 \times 60}}{{20}}} \right) \approx 138.6\,C\]
