Differential Equations

Second Order Equations

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Oscillations in Electrical Circuits

Solved Problems

Example 1.

An electrical circuit consists of a series-connected resistor \(R = 100\;\text{ohms}\) and a coil with inductance \(L = 50\;\text{H}.\) At time \(t = 0\) a \(DC\) source with the voltage \({V_0} = 200\;\text{V}\) is connected. Find:

  • the current change \(I\left( t \right)\) in the circuit;
  • the voltage change across the resistor \({V_R}\left( t \right)\) and the inductor \({V_L}\left( t \right)\).

Solution.

The series \(RL-\)circuit is described by the differential equation

\[L\frac{{dI}}{{dt}} + RI = {V_0}.\]

In accordance with the general theory, the solution of this equation is the sum of the general solution of the homogeneous equation \({I_0}\) and a particular solution of the nonhomogeneous equation \({I_1}:\) \(I = {I_0} + {I_1}.\) The general solution of the homogeneous equation

\[L\frac{{dI}}{{dt}} + RI = 0\]

is expressed as

\[{I_0}\left( t \right) = A{e^{ - \frac{R}{L}t}},\]

where \(A\) is the constant of integration.

The solution of the nonhomogeneous equation \({I_1}\) corresponds to the steady state in which the current in the circuit is determined only by the ohmic resistance \(R:\) \({I_1} = \frac{{{V_0}}}{R}.\) Then the total current varies according to the law

\[I\left( t \right) = {I_0} + {I_1} = A{e^{ - \frac{R}{L}t}} + \frac{{{V_0}}}{R}.\]

The constant \(A\) is determined from the initial condition \(I\left( {t = 0} \right) = 0.\) Consequently,

\[0 = A{e^{ - \frac{R}{L} \cdot 0}} + \frac{{{V_0}}}{R},\;\; \Rightarrow A = - \frac{{{V_0}}}{R}.\]

So, after the circuit is closed, the current will vary according to the law

\[I\left( t \right) = - \frac{{{V_0}}}{R}{e^{ - \frac{R}{L}t}} + \frac{{{V_0}}}{R} = \frac{{{V_0}}}{R}\left( {1 - {e^{ - \frac{R}{L}t}}} \right) = \frac{{200}}{{100}}\left( {1 - {e^{ - \frac{{100}}{{50}}t}}} \right) = 2\left( {1 - {e^{ - 2t}}} \right)\;\left[ \text{A} \right].\]

The graph \(I\left( t \right)\) is shown in Figure \(6.\)

Current change in RL-circuit
Figure 6.

The voltages \({V_R}\) across the resistor and \({V_L}\) ascross the inductor are determined by the following formulas:

\[{V_R}\left( t \right) = I\left( t \right)R = {V_0}\left( {1 - {e^{ - \frac{R}{L}t}}} \right) = 200\left( {1 - {e^{ - 2t}}} \right)\;\left[ \text{V} \right],\]
\[{V_L}\left( t \right) = L\frac{{dI\left( t \right)}}{{dt}} = \frac{{L{V_0}}}{R}\frac{d}{{dt}}\left( {1 - {e^{ - \frac{R}{L}t}}} \right) = \frac{{\cancel{L}{V_0}}}{\cancel{R}} \cdot \frac{\cancel{R}}{\cancel{L}}{e^{ - \frac{R}{L}t}} = {V_0}{e^{ - \frac{R}{L}t}} = 200{e^{ - 2t}}\;\left[ \text{V} \right].\]

The graphs of the functions \({V_R}\left( t \right)\) and \({V_L}\left( t \right)\) are shown in Figure \(7.\)

Voltage change on the resistance and inductance in RL-circuit
Figure 7.

Example 2.

An electrical circuit consists of a series-connected resistor \(R = 100\;\text{ohms}\) and a capacitor \(C = 0.01\;\mu\text{F}.\) At the initial moment a \(DC\) source with the voltage \({V_0} = 200\;\text{V}\) is connected to the circuit. Find:

  • the current change \(I\left( t \right)\) in the circuit;
  • the voltage change across the resistor \({V_R}\left( t \right)\) and the capacitor \({V_C}\left( t \right)\).

Solution.

This problem is similar to the previous and differs from it only by the type of electrical circuit. In this problem we consider an \(RC-\)circuit.

According to Kirchhoff's current law \(\left(KCL\right)\)

\[{V_R}\left( t \right) + {V_C}\left( t \right) = {V_0},\]

where the voltage across the resistor is given by

\[{V_R}\left( t \right) = I\left( t \right)R = RC\frac{{d{V_C}}}{{dt}}.\]

As a result, we obtain the following differential equation to describe the transition process in the \(RC\)-circuit:

\[RC\frac{{d{V_C}}}{{dt}} + {V_C} = {V_0}.\]

The solution of this equation is the sum of the general solution \({V_h}\) of the homogeneous equation and a particular solution \({V_1}\) of the nonhomogeneous equation. The homogeneous equation has the general solution \({V_h}\) in the form

\[RC\frac{{d{V_C}}}{{dt}} + {V_C} = 0,\;\; \Rightarrow \frac{{d{V_C}}}{{dt}} = - \frac{1}{{RC}}{V_C},\;\; \Rightarrow \int {\frac{{d{V_C}}}{{{V_C}}}} = - \frac{1}{{RC}}\int {dt} ,\;\; \Rightarrow \ln {V_C} = - \frac{t}{{RC}},\;\; \Rightarrow {V_h} = A{e^{ - \frac{t}{{RC}}}},\]

where \(A\) is the constant of integration, depending on initial conditions.

A particular solution of the nonhomogeneous equation corresponds to the steady state in which \(\frac{{d{V_C}}}{{dt}} = 0.\) Then the voltage across the resistor will be zero and all the voltage is applied to the capacitor, that is, \({V_C} = {V_0}.\) Thus, the voltage change on the capacitor is described by

\[{V_C}\left( t \right) = A{e^{ - \frac{t}{{RC}}}} + {V_0}.\]

Given the initial condition \({V_C}\left( {t = 0} \right) = 0,\) we find the constant \(A:\)

\[0 = A \cdot 1 + {V_0},\;\; \Rightarrow A = - {V_0}.\]

Consequently, the voltage change on the capacitor will look like this:

\[{V_C}\left( t \right) = - {V_0}{e^{ - \frac{t}{{RC}}}} + {V_0} = {V_0}\left( {1 - {e^{ - \frac{t}{{RC}}}}} \right) = 200\left( {1 - {e^{ - t}}} \right)\;\left[ \text{V} \right].\]

The voltage across the resistor is determined by the formula

\[{V_R}\left( t \right) = RC\frac{{d{V_C}}}{{dt}} = RC{V_0}\frac{d}{{dt}}\left( {1 - {e^{ - \frac{t}{{RC}}}}} \right) = \cancel{RC}{V_0} \cdot \frac{1}{\cancel{RC}}{e^{ - \frac{t}{{RC}}}} = {V_0}{e^{ - \frac{t}{{RC}}}} = 200{e^{ - t}}\;\left[ \text{V} \right].\]

The current in the \(RC-\)circuit will vary according to the law

\[I\left( t \right) = \frac{{{V_R}\left( t \right)}}{R} = \frac{{{V_0}}}{R}{e^{ - \frac{t}{{RC}}}} = \frac{{200}}{{100}}{e^{ - t}} = 2{e^{ - t}}\;\left[ \text{A} \right].\]

Graphs of the voltages \({V_C}\left( t \right),\) \({V_R}\left( t \right)\) and current \(I\left( t \right)\) are shown in Figures \(8\) and \(9.\)

Voltage change on the resistance and capacitor in RC-circuit
Figure 8.
Current change in RC-circuit
Figure 9.

Example 3.

An electrical circuit consists of a series-connected resistor \(R = 1\;\text{ohms},\) a coil with inductance \(L = 0.25\;\text{H}\) and a capacitor \(C = 1\;\mu\text{F}.\) How many oscillations will it make before the amplitude of the current is reduced by a factor of \(e?\)

Solution.

In this circuit, damped oscillations will occur with a frequency

\[\omega = \sqrt {\frac{1}{{LC}} - \frac{{{R^2}}}{{4{L^2}}}} .\]

The amplitude of the oscillations will decrease according to the law

\[A\left( t \right) = {A_0}{e^{ - \frac{R}{{2L}} t}}.\]

Suppose that \(N\) complete oscillations occurred for time \(t:\)

\[t = NT = \frac{{2\pi N}}{\omega } = \frac{{2\pi N}}{{\sqrt {\frac{1}{{LC}} - \frac{{{R^2}}}{{4{L^2}}}} }}.\]

If the amplitude decreased by \(e\) times, then one can write the following equation:

\[- \frac{R}{{2L}}t = \frac{R}{{2L}} \cdot \frac{{2\pi N}}{{\sqrt {\frac{1}{{LC}} - \frac{{{R^2}}}{{4{L^2}}}} }} = - 1.\]

Hence, we find the number of oscillations \(N:\)

\[N = \frac{L}{{\pi R}}\sqrt {\frac{1}{{LC}} - \frac{{{R^2}}}{{4{L^2}}}} = \frac{1}{\pi }\sqrt {\frac{{{L^\cancel{2}}}}{{{R^2}\cancel{L}C}} - \frac{{\cancel{L^2}\cancel{R^2}}}{{4\cancel{R^2}\cancel{L^2}}}} = \frac{1}{\pi }\sqrt {\frac{L}{{{R^2}C}} - \frac{1}{4}} = \frac{1}{{2\pi }}\sqrt {\frac{{4L}}{{{R^2}C}} - 1} = \frac{1}{{2\pi }}\sqrt {\frac{{4 \cdot 0.25}}{{{1^2} \cdot {{10}^{ - 6}}}} - 1} \approx \frac{{1000}}{{2\pi }} \approx 159.\]

Example 4.

An \(AC\) source with amplitude \({E_0} = 128\;\text{V}\) and frequency \(\omega = 250\;\text{Hz}\) is connected to a series circuit consisting of a resistance \(R = 100\;\text{ohms},\) a coil with inductance \(L = 0.4\;\text{H}\) and a capacitor \(C = 200\;\mu\text{F}.\) Find:

  • the current amplitude in the circuit;
  • the voltage amplitude on the capacitor.

Solution.

The current oscillations in the steady state occur with the amplitude

\[{I_0} = \frac{{{E_0}}}{{\sqrt {{R^2} + {{\left( {\omega L - \frac{1}{{\omega C}}} \right)}^2}} }} = \frac{{128}}{{\sqrt {{{10}^4} + {{\left( {250 \cdot 0.4 - \frac{1}{{250 \cdot 0.2 \cdot {{10}^{ - 3}}}}} \right)}^2}} }} = \frac{{128}}{{\sqrt {{{10}^4} + {{\left( {100 - 20} \right)}^2}} }} = \frac{{128}}{{\sqrt {16400} }} \approx {1\;\left[ \text{A} \right].}\]

The amplitude of the voltage oscillations on the capacitor will be equal to

\[{V_C} = \frac{{{q_0}}}{C} = \frac{{{I_0}}}{{\omega C}} = \frac{1}{{250 \cdot 0,2 \cdot {{10}^{ - 3}}}} = 20\;\left[ \text{V} \right].\]
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