5.1 RL circuits

Recall that a changing current through an inductor induces a voltage that opposes the change, \(\mathcal E_L = -L\,dI/dt\) (3.6), where the inductance \(L\) depends only on geometry.

This is the same equation as for the RC circuit, with \(Q\leftrightarrow I\). Recall (2.2), the loop rule \(V - IR - Q/C = 0\) with \(I = dQ/dt\): \[V - \frac{Q}{C} - R\,\frac{dQ}{dt} = 0, \qquad Q(t) = VC\left(1 - e^{-t/RC}\right).\] Under the replacements \[Q \to I, \qquad R \to L, \qquad C = \frac{1}{R}\] this becomes (5.1), so \[\keyeq{\displaystyle I(t) = \frac{V}{R}\left(1 - e^{-(R/L)\,t}\right)}\]

Connection with the oscillations chapter

We can also solve (5.1) with the strategy of the Oscillations chapter. The homogeneous equation \(L\,dI/dt + RI = 0\) is solved by guessing \(I = I_0\,e^{\lambda t}\), which gives \(\lambda = -R/L\). A constant \(I = V/R\) solves the full equation, so \(I(t) = V/R + I_0\,e^{-(R/L)t}\), and \(I(0) = 0\) fixes \(I_0 = -V/R\): the same result. This split into a homogeneous and a particular solution reappears for the driven RLC circuit below.

5.2 LC circuits

Compare (4.1): \(Q\) plays the role of \(x\) and \(1/LC\) the role of \(k/m\). By (4.4), \[\keyeq{\displaystyle Q(t) = Q_0\cos(\omega t + \phi)}\;, \qquad \keyeq{\displaystyle \omega = \frac{1}{\sqrt{LC}}}\] where \(\omega\) is the natural angular frequency of the circuit.

5.3 RLC circuits

Connection with the damped oscillator

Recall (4.5), \(m\,\ddot x + b\,\dot x + kx = 0\). Equation (5.3) is the same equation with \[m = 1, \qquad b = \frac{R}{L}, \qquad k = \frac{1}{LC}, \qquad x \leftrightarrow Q .\] (Multiplying (5.3) by \(L\) gives the equivalent dictionary: inductance \(\leftrightarrow\) mass, resistance \(\leftrightarrow\) damping coefficient \(b\), \(1/C \leftrightarrow\) spring constant \(k\).) The decay time \(\tau = 2m/b\) becomes \(2L/R\), and \(\omega_0 = \sqrt{k/m}\) becomes the LC frequency \(1/\sqrt{LC}\).

The solution of the damped oscillator then gives, in the underdamped case \(1/LC > (R/2L)^2\), \[\keyeq{\displaystyle Q(t) = Q_0\,e^{-(R/2L)\,t}\cos(\omega t + \phi)}\] with \[\keyeq{\displaystyle \omega = \sqrt{\frac{1}{LC} - \left(\frac{R}{2L}\right)^2}}\;, \qquad T = \frac{2\pi}{\omega} .\]

5.4 Driven RLC circuit

Strategy: homogeneous plus particular solution

Solve for \(Q = Q_h + Q_p\) (the general solution). The homogeneous part \(Q_h\) solves the equation without the source, \(L\ddot Q_h + R\dot Q_h + Q_h/C = 0\), so by the previous section \[Q_h(t) = Q_0\,e^{-(R/2L)\,t}\cos(\omega' t + \phi), \qquad \omega' = \sqrt{\frac{1}{LC} - \left(\frac{R}{2L}\right)^2}.\] This decays exponentially, so at late times only the particular solution \(Q_p\) remains.

Finding \(Q_p\)

Guess a response at the driving frequency, \[Q_p = \frac{A}{\omega}\cos\omega t + \frac{B}{\omega}\sin\omega t .\] Then the current we are looking for is \[I_p = \frac{dQ_p}{dt} = -A\sin\omega t + B\cos\omega t, \qquad \frac{d^2Q_p}{dt^2} = -A\omega\cos\omega t - \omega B\sin\omega t .\] Plugging into (5.4) gives \[\begin{align*} V_0\sin\omega t &= L\left(-A\omega\cos\omega t - B\omega\sin\omega t\right) \\ &\quad + R\left(-A\sin\omega t + B\cos\omega t\right) \\ &\quad + \frac{1}{\omega C}\left(A\cos\omega t + B\sin\omega t\right) \end{align*}\] Define \[R_{LC} := \omega L - \frac{1}{\omega C} .\] Matching the terms in \(\cos\omega t\): \[-L\omega A + RB + \frac{1}{\omega C}A = 0 \;\Longleftrightarrow\; -R_{LC}A + RB = 0 \;\Longrightarrow\; \keyeq{\displaystyle A = \frac{R}{R_{LC}}\,B}\] and in \(\sin\omega t\): \[-L\omega B - RA + \frac{1}{\omega C}B = V_0 \;\Longleftrightarrow\; -R_{LC}B - RA = V_0 \;\Longrightarrow\; \left(R_{LC} + \frac{R^2}{R_{LC}}\right)B = -V_0 .\] Hence \[\keyeq{\displaystyle B = -V_0\,\frac{R_{LC}}{R_{LC}^2 + R^2}} \qquad \keyeq{\displaystyle A = -V_0\,\frac{R}{R_{LC}^2 + R^2}}\]

Amplitude and phase of the current

The particular current \(I_p = B\cos\omega t - A\sin\omega t\) has the form \(C\cos\omega t + D\sin\omega t\) of the Oscillations chapter, so by (4.4) \[I_p(t) = I_0\cos(\omega t + \phi), \qquad I_0 = \sqrt{A^2 + B^2}, \qquad \tan\phi = \frac{A}{B}.\]

Substituting \(A\) and \(B\), \[\keyeq{\displaystyle I_0 = V_0\sqrt{\frac{R_{LC}^2 + R^2}{\left(R_{LC}^2 + R^2\right)^2}} = \frac{V_0}{\sqrt{R_{LC}^2 + R^2}}}\]

Equivalently \(I_0 = V_0/R_{\mathrm{eff}}\) with the effective resistance \[R_{\mathrm{eff}} = \sqrt{\left(\omega L - \frac{1}{\omega C}\right)^2 + R^2} .\]

Resonance

The amplitude \(I_0\) is largest when \(R_{LC} = \omega L - 1/(\omega C) = 0\), that is at \[\keyeq{\displaystyle \omega_0 = \frac{1}{\sqrt{LC}}}\] the natural frequency of the LC circuit, where \(I_0 = V_0/R\). This is resonance!

Quality factor

The quality factor \(\mathsf{Q}\) (not to be confused with the charge \(Q\)) measures how narrow the resonance is, \[\keyeq{\displaystyle \mathsf{Q} = \frac{\omega_0}{\Delta\omega}}\] where \(\Delta\omega\) is the width of the peak between the points where the dissipated power (\(\propto I_0^2\)) has dropped to half its maximum, i.e. where \(I_0 = I_{\max}/\sqrt2\), equivalently \(|R_{LC}| = R\). When the peak is narrow (small \(R\)), we can expand around \(\omega_0\): \[R_{LC} = \frac{L\left(\omega^2 - \omega_0^2\right)}{\omega} \approx 2L\,(\omega - \omega_0),\] so \(|R_{LC}| = R\) at \(|\omega - \omega_0| = R/2L\), a full width \(\Delta\omega \approx R/L\). Therefore \[\keyeq{\displaystyle \mathsf{Q} \approx \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}}\] Using \(\tau = 2L/R\) from the damped oscillator, this is also \(\mathsf{Q}\approx\omega_0\tau/2\): a large \(\mathsf Q\) means weak damping and a sharp resonance.

Problems

Problems are taken from the sources listed under Sources and Attribution. Numbers refer to the end-of-chapter problems; figures have been redrawn.

OpenStax University Physics Vol. 2, Chapter 14: Inductance

OpenStax University Physics Vol. 2, Chapter 15: Alternating-Current Circuits