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PHY294H - Lecture 30

RL circuits, DC source

Kirchhoff's loop law for a circuit containing a DC voltage $\epsilon$, a resistor R and an inductor L is,

\begin{displaymath}\epsilon - IR - L{dI \over dt} = 0
\end{displaymath} (1)

The solution to this equation is,

\begin{displaymath}I(t) = a + be^{-t/\tau}
\end{displaymath} (2)


\begin{displaymath}\tau = L/R
\end{displaymath} (3)

The constants a and bdepend on the initial conditions.

Initial current is zero

In this case, we imagine that a switch is in the circuit and that the switch is open for t<0. We thus have I(0) = 0. This implies that

a + b = 0 (4)

Since there is no current at time t=0, the voltage across the resistor is zero at t=0. Therefor the entire voltage is dropped across the inductor. We thus have,

\begin{displaymath}V_L(0) = L{dI\over dt}\vert _{t=0} = \epsilon = -{b L\over \tau} = -bR
\end{displaymath} (5)

From this equation we find,

\begin{displaymath}b = -{\epsilon\over R} = \ \ \ ; \ \ \ a = -b = {\epsilon\over R}
\end{displaymath} (6)

The current in the circuit is then,

\begin{displaymath}I(t) = {\epsilon\over R} (1-e^{-t/\tau})
\end{displaymath} (7)

At long times($t>>\tau$), the inductance plays no role in this DC circuit.

Initial current is I0, no applied voltage

In this case we have,

I(0) = I0 = a + b (8)

We also have

\begin{displaymath}V_R = I_0R = -V_L = -L{dI\over dt} = bRe^{-t/\tau} = bR
\end{displaymath} (9)

We then have,

\begin{displaymath}b = I_0\ \ \ ;\ \ \ a=0
\end{displaymath} (10)

so that,

\begin{displaymath}I(t) = I_0e^{-t/\tau}
\end{displaymath} (11)

An LC circuit - an oscillator

A circuit with an inductor and a capacitor acts as an oscillator. There is no dissipation in this circuit, provided the capacitor and the inductor are ideal.

We shall consider a case in which the capacitor has charge Q0 at t=0 and that at this time there is no current flowing in the circuit. The capacitor at time t=0 stores energy Q02/C, while the inductor stores no energy at t=0. At time t=0, we then have,

\begin{displaymath}Q(0) = Q_0\ \ \ ;\ \ \ I(0) = 0
\end{displaymath} (12)

As time goes on, energy is exhanged between the capacitor and the inductor. This is very similar to the exchange between kinetic and potential energy in a spring-mass system or in a pendulum.

Kirchhoff's loop law for the LC ciruit is,

\begin{displaymath}L{dI\over dt} + {Q\over C} = 0
\end{displaymath} (13)

Since I=dQ/dt, this equation is equivalent to,

\begin{displaymath}L {d^2Q\over dt^2} + {Q\over C} = 0
\end{displaymath} (14)

This is the equation for simple harmonic motion and has the general solution,

\begin{displaymath}Q(t) = A Cos(\omega t + \phi)
\end{displaymath} (15)

where A and $\phi$ are constants which we may choose to satisfy the initial conditions. In our example, the initial conditions are, Q(0) = Q0 and I(0) = 0, which are satisfied by,

\begin{displaymath}Q(t) = Q_0 cos(\omega t)
\end{displaymath} (16)

LRC circuit

We are just adding a resistor to the problem we solved above. Kirchhoff's loop law gives

\begin{displaymath}L{dI\over dt} + IR+{Q\over C} = 0
\end{displaymath} (17)

Using I=dQ/dt we find,

\begin{displaymath}L{d^2Q \over dt^2} + R{dQ\over dt} + {Q\over C} = 0
\end{displaymath} (18)

The general solution to this equation is,

\begin{displaymath}Q(t) = Ae^{-\alpha t} cos(\omega' t + \phi)
\end{displaymath} (19)

where A and $\phi$ are constants which are chosen to fit the initial conditions. $\omega'$ and $\alpha$ are found by subsituting Eq. (19) into equation (18) and ensuring that Eq. (19) is a solution for all time. To carry out this calculations, we need to take derivatives of the charge with the respect to time. The first derivative is,

\begin{displaymath}{dQ\over dt} = -Aw'e^{-\alpha t}sin(w't+\phi)
- A\alpha e^{-\alpha t} cos(w't + \phi)
\end{displaymath} (20)


\begin{displaymath}{d^2Q\over dt^2} = -Aw'^2e^{-\alpha t}cos(w't+\phi)
+ 2Aw'\a...
...\alpha t} sin(w't+\phi)
+ A \alpha^2e^{-\alpha t}cos(w't+\phi)
\end{displaymath} (21)

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Phil Duxbury