Taking the cell away
Let a steady current \(I_0 = E/R\) flow in an LR circuit. Now throw a switch that takes the cell out and closes the loop with a plain wire. With no cell, nothing drives the current, yet the current cannot stop at once. The inductor's back emf now pushes the current along, and the current dies away gradually. This is the decay of current.
\[ 0 = IR + L\dfrac{dI}{dt} \]
Why. the loop equation with no cell.
\[ \dfrac{dI}{I} = -\dfrac{R}{L}\,dt \]
Why. separate the current from the time.
\[ \ln\dfrac{I}{I_0} = -\dfrac{Rt}{L} \]
Why. integrate from \(I_0\) at \(t = 0\) to \(I\) at \(t\).
\[ I = I_0\,e^{-t/\tau}, \quad \tau = \dfrac{L}{R} \]
Read it. take the exponential of both sides.
Follow Fig 6.1. The current keeps flowing the same way round, through the same resistor, and falls smoothly. Now the inductor is driving the current, so it acts as the source of the loop. Its polarity has reversed: the end where the current enters is now the negative one. With \(8\,\Omega\) in the loop, the time constant is \(2/8 = 0.25\,\text{s}\), and the decay is twice as fast.
A \(2\,\text{H}\) inductor carries \(3\,\text{A}\) through a \(4\,\Omega\) resistor when the cell is replaced by a wire. Find (a) the current at \(0.5\,\text{s}\) and at \(1\,\text{s}\), (b) the time for the current to fall to \(10\,\%\), and (c) the total heat produced.
(a) \(\tau = 2/4 = 0.5\,\text{s}\). At \(0.5\,\text{s} = \tau\), \(I = 3e^{-1} = 3\times0.37 = 1.1\,\text{A}\). At \(1\,\text{s} = 2\tau\), \(I = 3e^{-2} = 0.41\,\text{A}\). With the workbook's rounding, \(e^{-2} = 0.14\), this is \(0.42\,\text{A}\).
(b) \(e^{-t/\tau} = 0.1\), so \(t = \tau\ln10 = 0.5\times2.3 = 1.15\,\text{s}\).
(c) The stored energy \(\tfrac12 LI_0^2 = \tfrac12\times2\times3^2 = 9\,\text{J}\) all becomes heat in the resistor.
Values used: \(e^{-1} = 0.37\), \(e^{-2} = 0.14\) and \(\ln10 = 2.3\), as the workbook rounds them; exactly, \(3e^{-2} = 0.406\,\text{A}\).
The heat equals the stored energy exactly. The resistor turns \(I^2R\) into heat at every instant, and the store falls by just that much. So a larger resistance spends the same \(9\,\text{J}\), only faster. With \(8\,\Omega\) the current is gone in half the time, but the heat is still \(9\,\text{J}\).
A \(0.4\,\text{H}\) coil carrying \(2\,\text{A}\) is closed through a total resistance of \(8\,\Omega\), with no cell. What is the current \(0.1\,\text{s}\) later, in amperes?
How much heat does the resistor of the last check produce while the current dies away completely, in joules?
A current decays in an LR circuit with time constant \(\tau\). Put these moments in order, earliest first.
Reading \(\tau\) off a graph
The decay curve starts steeply and then flattens. Its slope at the start is \(dI/dt = -I_0/\tau\). A straight line with that slope, drawn from \(I_0\), would reach zero after exactly \(\tau\). So draw the tangent at \(t = 0\), and where it meets the time axis, read off the time constant. It is the dashed line in Fig 6.1.
\[ \dfrac{dI}{dt} = -\dfrac{I_0}{\tau}\,e^{-t/\tau} \]
Why. differentiate \(I = I_0e^{-t/\tau}\).
\[ \left.\dfrac{dI}{dt}\right|_{t = 0} = -\dfrac{I_0}{\tau} \]
Why. put \(t = 0\).
\[ I = I_0 - \dfrac{I_0}{\tau}\,t = 0 \quad\Rightarrow\quad t = \tau \]
Read it. the tangent line starts at \(I_0\) with this slope and reaches zero at \(t = \tau\).
In the example the tangent starts at \(3\,\text{A}\) with a slope of \(-3/0.5 = -6\,\text{A/s}\). It meets the axis at \(0.5\,\text{s}\). The growth curve has the same property, as the last lesson showed: its tangent at \(t = 0\) reaches the final current at \(t = \tau\).
On the graph of a decaying current, the tangent at \(t = 0\) meets the time axis at \(40\,\text{ms}\). The circuit has \(R = 5\,\Omega\). What is its inductance?
Opening the switch on the two-branch circuit
Go back to the cell, \(R_1\), and the inductor in parallel with \(R_2\), all at \(4\,\Omega\), \(12\,\text{V}\) and \(2\,\text{H}\). After the switch has been closed a long time, the inductor carries \(3\,\text{A}\) and \(R_2\) carries nothing. Now the switch is opened.
Just after the switch opens, what current flows through \(R_2\)?
The inductor's current cannot change in an instant, so its \(3\,\text{A}\) goes on flowing down through it. With the cell's branch open, the only way back is up through \(R_2\). So \(R_2\) suddenly carries \(3\,\text{A}\), opposite to its current while the switch was closing. For an instant it has \(3\times4 = 12\,\text{V}\) across it. The current then dies away round the loop of \(L\) and \(R_2\), with \(\tau = L/R_2 = 2/4 = 0.5\,\text{s}\).
Without \(R_2\) the inductor's current would have no path at all. It would then be forced to zero almost at once, and the back emf would strike a spark across the switch. The parallel path lets the current, and the stored energy, die away gently.
A relay is a switch worked by a small electromagnet, and one clicks inside a home inverter when the mains fails. When the transistor feeding its coil switches off, the coil's current has nowhere to go. Its back emf could reach hundreds of volts and destroy the transistor. So a diode is wired across the coil, pointing so that it carries no current while the coil is fed. At switch-off it gives the coil's current a parallel path, and the current dies away through it, as through \(R_2\) above.
The current changes smoothly, the voltage need not
Every result of these two lessons rests on one rule. The current through an inductor cannot jump. A jump would mean an infinite \(dI/dt\), and so an infinite back emf \(L\,dI/dt\), which no source can supply. The stored energy \(\tfrac12 LI^2\) cannot jump either, for the same reason.
The voltage across an inductor has no such limit. At switch-on in the last lesson, \(V_L\) jumped from \(0\) to \(E\). In the circuit above, the voltage across the inductor jumped to \(12\,\text{V}\) with its polarity reversed. When you meet a switched circuit, first ask what current each inductor carried just before. The same currents flow just after.
A switch operates in a circuit that contains an inductor. Which quantity cannot change suddenly at that instant?
- Letting the current stop the instant the cell is removed.The inductor keeps it flowing in the same direction, and it dies away as \(I_0e^{-t/\tau}\).
- Keeping the growth formula for a decay.With no source in the loop, \(I = I_0e^{-t/\tau}\); it starts at \(I_0\) and falls towards zero.
- Thinking a larger resistance releases less heat.The heat is always \(\tfrac12 LI_0^2\). A larger resistance releases it faster.
- Assuming a resistor's current keeps its direction when a switch opens.Only the inductor's current is fixed. A resistor in parallel with it may carry that current the other way.