# Differential Equations

2nd Order Equations# Oscillations in Electrical Circuits

Theory

Problems 1-4

### Differential Equations of RLC-Circuits

Electric oscillations can be excited in a circuit containing resistance \(R\), inductance \(L\) and capacitance \(C\). In terms of topology, two types of circuits are often considered: series \(RLC\)-circuit (Figure 1) and parallel \(RLC\)-circuit (Figure \(2\)).

We derive the differential equation describing the current change in a series \(RLC\) circuit.

The voltages \({V_R},{V_C},{V_L},\) respectively, on the resistor \(R,\) capacitor \(C\) and inductor \(L\) are given by

{{V_R}\left( t \right) = RI\left( t \right),\;\;\;}\kern-0.3pt

{{V_C}\left( t \right) = \frac{1}{C}\int\limits_0^t {I\left( \tau \right)d\tau } ,\;\;\;}\kern-0.3pt

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

\]

It follows from the Kirchhoff’s voltage law \(\left(KVL\right)\) that

Figure 1.

Figure 2.

where \(E\left( t \right)\) is the electromotive force (emf) of the power supply.

In the case of constant emf \(E,\) we obtain the following differential equation after substituting the expressions for \({V_R},\) \({V_C},\)\({V_L}\) and differentiation:

If we denote \(2\beta = {\large\frac{R}{L}\normalsize},\) \(\omega _0^2 = {\large\frac{1}{{LC}}\normalsize},\) the equation can be written as

This differential equation coincides with the equation describing the damped oscillations of a mass on a spring. Hence, damped oscillations can also occur in series *RLC*-circuits with certain values of the parameters.

Now consider the parallel \(RLC\)-circuit and derive a similar differential equation for it.

By the Kirchhoff’s current law \(\left(KCL\right)\), the total current is equal to the sum of currents through a resistor \(R,\) inductor \(L\) and capacitor \(C\) (Figure \(2\)):

Given that

{{I_R} = \frac{V}{R},\;\;\;}\kern-0.3pt

{{I_L} = \frac{1}{L}\int\limits_0^t {Vd\tau } ,\;\;\;}\kern-0.3pt

{{I_C} = C\frac{{dV}}{{dt}},}

\]

for the case of constant total current \(I\left( t \right) = {I_0},\) we obtain the following differential equation of the second order with respect to the variable \(V:\)

{{\frac{V}{R} + \frac{1}{L}\int\limits_0^t {Vd\tau } }+{ C\frac{{dV}}{{dt}} = {I_0},\;\;}}\Rightarrow

{{C\frac{{{d^2}V}}{{d{t^2}}} + \frac{1}{R}\frac{{dV}}{{dt}} }+{ \frac{1}{L}V = 0.}}

\]

As one can see, we again have the equation describing the damped oscillations. Thus, the oscillatory mode can also occur in parallel \(RLC\)-circuits.

### Resonant Circuit. Thomson Formula

In the simplest case, when the ohmic resistance is zero \(\left(R = 0\right)\) and the source of emf is removed \(\left(E = 0\right)\), the resonant circuit consists only of a capacitor \(C\) and inductor \(L,\) and is described by the differential equation

In this circuit there will be undamped electrical oscillations with a period

This formula is called the Thomson formula in honor of British physicist William Thomson \(\left(1824-1907\right)\), who derived it theoretically in \(1853.\)

### Damped Oscillations in Series RLC-Circuit

The second order differential equation describing the damped oscillations in a series \(RLC\)-circuit we got above can be written as

The corresponding characteristic equation has the form

Its roots are calculated by the formulas:

{{\lambda _{1,2}} }={ \frac{{ – \frac{R}{L} \pm \sqrt {\frac{{{R^2}}}{{{L^2}}} – \frac{4}{{LC}}} }}{2} }

= { – \frac{R}{{2L}} \pm \sqrt {{{\left( {\frac{R}{{2L}}} \right)}^2} – \frac{1}{{LC}}} }

= { – \beta \pm \sqrt {{\beta ^2} – \omega _0^2} ,}

\]

where the value of \(\beta = {\large\frac{R}{{2L}}\normalsize}\) is called the damping coefficient, and \({\omega_0}\) is the resonant frequency of the circuit.

Depending on the values of \(R, L, C\) there may be three options.

Figure 3.

### Case 1. Overdamping: \({R^2} \gt {\large\frac{{4L}}{C}\normalsize}\)

In this case, both roots of the characteristic equation \({\lambda_1}\) and \({\lambda_2}\) and real, distinct and negative. The general solution of the differential equation is given by

In this mode, the current decreases monotonically approaching zero (Figure \(3\text{).}\)

### Case 2. Critical Damping: \({R^2} = {\large\frac{{4L}}{C}\normalsize}\)

This mode can be called boundary or critical. Here, both roots of the characteristic equation are equal, real and negative. The general solution is expressed by the function

{I\left( t \right) = \left( {{C_1}t + {C_2}} \right){e^{ – \beta t}} }

= {\left( {{C_1}t + {C_2}} \right){e^{ – {\large\frac{R}{{2L}}\normalsize} t}}.}

\]

At the beginning of the process, the current may even increase, but then it quickly decreases exponentially.

### Case 3. Underdamping: \({R^2} \lt {\large\frac{{4L}}{C}\normalsize}\)

In this case, the roots of the characteristic equation are complex conjugate, which leads to damped oscillations in the circuit. The change of current is given by

where the value of \(\beta = {\large\frac{R}{{2L}}\normalsize}\) is as above the damping factor, \(\omega =\) \( \sqrt {{\large\frac{1}{{LC}}\normalsize} – {{\left( {\large\frac{R}{{2L}}\normalsize} \right)}^2}} \) is the frequency of oscillation, \(A, B\) are constants of integration, depending on initial conditions. Note that the frequency \(\omega\) of damped oscillations is less than the resonant frequency \({\omega_0}\) of the circuit. The typical shape of the curve \(I\left( t \right)\) in this mode is also shown in Figure \(3\) above.

### Forced Oscillations and Resonance

If the resonant circuit includes a generator with periodically varying emf, the forced oscillations arise in the system. If the emf \(E\) of the source varies according to the law

then the differential equation of forced oscillations in series \(RLC\)-circuit can be written as

{{\frac{{{d^2}q\left( t \right)}}{{d{t^2}}} + \frac{R}{L}\frac{{dq\left( t \right)}}{{dt}} }+{ \frac{1}{{LC}}q\left( t \right) }={ \frac{1}{L}{E_0}\cos \omega t\;\;}}\kern-0.3pt

{{\text{or}\;\;\frac{{{d^2}q}}{{d{t^2}}} + 2\beta \frac{{dq}}{{dt}} + \omega _0^2q }={ \frac{{{E_0}}}{L}\cos \omega t,}}

\]

where \(q\) the charge of the capacitor, \(2\beta = \frac{R}{L},\) \(\omega _0^2 = \frac{1}{{LC}}.\)

This equation is analogous to the equation of forced oscillations of a spring pendulum, discussed on the page Mechanical Oscillations. Its general solution is the sum of two components: the general solution of the associated homogeneous equation and a particular solution of the nonhomogeneous equation. The first component describes the decaying transient process, after which the behavior of the system depends only on the external driving force. The forced oscillations will occur according to the law

{q\left( t \right) \text{ = }}\kern0pt

{{\frac{{{E_0}}}{{L\sqrt {{{\left( {\omega _0^2 – {\omega ^2}} \right)}^2} + 4{\beta ^2}{\omega ^2}} }} \cdot}}\kern0pt{{ \cos \left( {\omega t + \varphi } \right) }}

= {{\frac{{{E_0}}}{{\omega \sqrt {{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}} }} \cdot}\kern0pt{ \cos \left( {\omega t + \varphi } \right),}}

\]

where the phase \(\varphi\) is determined by the formula

{\varphi = \arctan \left( { – \frac{{2\beta \omega }}{{\omega _0^2 – {\omega ^2}}}} \right) }

= {\arctan \frac{R}{{\omega L – \frac{1}{{\omega C}}}}.}

\]

Knowing the change of the charge \(q\left( t \right),\) it is easy to find the change of the current \(I\left( t \right):\)

{I\left( t \right) = \frac{{dq\left( t \right)}}{{dt}} }

= {{ – \frac{{{E_0}}}{{\sqrt {{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}} }} \cdot}\kern0pt{ \sin\left( {\omega t + \varphi } \right) }}

= {{\frac{{{E_0}}}{{\sqrt {{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}} }} \cdot}\kern0pt{ \cos\left( {\omega t – \theta } \right),}}

\]

where we have introduced the angle \(\theta\) such that \(\theta = – \left( {\varphi + \frac{\pi }{2}} \right).\) The angle \(\theta\) indicates the phase shift of the current oscillations \(I\left( t \right)\) with respect to oscillations in the supply voltage \(E\left( t \right) = {E_0}\cos \omega t.\)

The amplitude of the current \({I_0}\) and the phase shift \(\theta\) are given by

{{{I_0} }={ \frac{{{E_0}}}{{\sqrt {{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}} }} }={ \frac{{{E_0}}}{Z},\;\;\;}}\kern-0.3pt

{\theta = \arctan \frac{{\omega L – \frac{1}{{\omega C}}}}{R}.}

\]

The quantity \(Z =\) \( \sqrt {{R^2} + {{\left( {\omega L – {\large\frac{1}{{\omega C}}\normalsize}} \right)}^2}} \) is called the impedance, or impedance of the circuit. It consists of an ohmic resistance \(R\) and a reactance \({\omega L – {\large\frac{1}{{\omega C}}}\normalsize}.\) Impedance of the resonant circuit in the complex form can be written as

We see from these formulas that the amplitude of steady-state oscillations of the current is maximum when

On this condition, resonance appears in the circuit. The resonant frequency \({\omega_0}\) is equal to the frequency of free oscillations in the circuit and does not depend on the resistance \(R.\)

We can transform the formula for the amplitude of the forced oscillations to get an explicit dependence on the frequency ratio \(\large\frac{\omega }{{{\omega _0}}}\normalsize,\) where \({\omega_0}\) is the resonant frequency. As a result, we obtain

{{I_0} = \frac{{{E_0}}}{{\sqrt {{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}} }} }

= {\frac{{\frac{{{E_0}}}{{{\omega _0}}}}}{{\frac{{\sqrt {{R^2} + {{\left( {\omega L – \frac{1}{{\omega C}}} \right)}^2}} }}{{{\omega _0}}}}} }

= {\frac{{\frac{{{E_0}}}{{{\omega _0}}}}}{{\sqrt {\frac{{{R^2}}}{{\omega _0^2}} + {{\left( {\frac{\omega }{{{\omega _0}}}L – \frac{1}{{\omega {\omega _0}C}}} \right)}^2}} }} }

= {\frac{{{E_0}\sqrt {LC} }}{{\sqrt {{R^2}LC + {{\left( {\frac{\omega }{{{\omega _0}}}L – \frac{1}{{\frac{\omega }{{{\omega _0}}}\frac{\cancel{C}}{{L\cancel{C}}}}}} \right)}^2}} }} }

= {\frac{{{E_0}\sqrt {LC} }}{{\sqrt {{R^2}LC + {{\left( {\frac{\omega }{{{\omega _0}}}L – \frac{L}{{\frac{\omega }{{{\omega _0}}}}}} \right)}^2}} }} }

= {\frac{{{E_0}\sqrt C }}{{\sqrt {{R^2}C + {{\left( {\frac{\omega }{{{\omega _0}}} – \frac{1}{{\frac{\omega }{{{\omega _0}}}}}} \right)}^2}} }}.}

\]

Dependencies of the current amplitude on the frequency ratio \(\large\frac{\omega }{{{\omega _0}}}\normalsize\) for different values of \(R\) and \(C\) are shown in Figures \(4\) and \(5.\) These graphs are built at \(E = 100\;\text{V},\) \(L = 1\;\text{mH},\) \(C = 10\;\mu\text{F}\) (Figure \(4\)), \(R = 10\;\text{ohms}\) (Figure \(5\)).

Figure 4.

Figure 5.

Resonance properties of a circuit are characterized by the quality factor \(Q,\) which is numerically equal to the ratio of the resonance frequency \({\omega_0}\) to the width \(\Delta\omega\) of the resonance curve at \(\large{\frac{1}{\sqrt 2}}\normalsize\) of the maximum value (Figure \(6\)).

The \(Q\) factor in a series \(RLC\) circuit is given by

Figure 6.

For a parallel \(RLC\) circuit, the \(Q\) factor is determined by the inverse expression:

## Solved Problems

Click on problem description to see solution.

### ✓ 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 \({V_L}\left( t \right)\).

### ✓ 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 \({V_C}\left( t \right)\).

### ✓ 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 \(e?\)

### ✓ 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.

Theory

Problems 1-4