The LR Circuit: Growth of Current

Switching on a coil

Join a cell of emf \(E\), a resistor \(R\) and an inductor \(L\) in series, and close the switch at \(t = 0\). Without the inductor the current would jump at once to \(E/R\). With it, the current cannot jump, because a sudden jump would need an enormous back emf. So the current grows smoothly, and the circuit is called an LR circuit.

Go round the loop by Kirchhoff's loop rule, from the chapter Circuits: Kirchhoff's Laws, Networks and RC Transients. The cell raises the potential by \(E\). The resistor drops it by \(IR\), and the inductor by its back emf \(L\,dI/dt\). The rises and drops round a loop add to zero.

The quantity \(\tau = L/R\) is the time constant of the circuit. It really is a time. A henry is a volt second per ampere, and an ohm is a volt per ampere, so their ratio is a second. After one time constant the current has reached \(1 - e^{-1} = 0.63\) of its final value. After \(2\tau\) it is \(0.86\), after \(3\tau\) it is \(0.95\), and after \(5\tau\) it is within one per cent.

Play with the figure above. A larger \(L\) stretches the growth over a longer time but leaves the final current alone. A larger \(R\) lowers the final current \(E/R\) and shortens the time constant. At the first instant the current is zero and grows at the rate \(E/L\), whatever the resistance. The dashed tangent at \(t = 0\) would reach the final current at exactly \(t = \tau\).

Values used: \(e^{-1} = 0.37\) and \(\ln2 = 0.69\), as the workbook rounds them; the exact current at \(\tau\) is \(1.896\,\text{A}\). The same steps serve any LR circuit. A coil of \(0.1\,\text{H}\) and \(2\,\Omega\) on \(6\,\text{V}\) has \(\tau = 0.05\,\text{s}\) and a final current of \(3\,\text{A}\), and it carries \(1.9\,\text{A}\) after \(0.05\,\text{s}\). A \(50\,\text{mH}\) coil with \(10\,\Omega\) has \(\tau = 5\,\text{ms}\). A \(10\,\text{V}\) cell on a \(0.5\,\text{H}\) coil of negligible resistance starts its current rising at \(E/L = 20\,\text{A/s}\).

Where the cell's voltage goes

At every instant the cell's \(E\) is shared between the resistor and the inductor. The resistor takes \(V_R = IR = E(1 - e^{-t/\tau})\), which rises with the current. The inductor takes \(V_L = L\,dI/dt = E\,e^{-t/\tau}\), which falls as the current levels off. Their sum is always \(E\). While the current grows, the end of the inductor where the current enters is the positive one.

At the instant of closing, \(V_L = E\): the whole emf of the cell sits across the inductor. The back emf exactly balances the cell, leaving nothing to drive a current through \(R\). That is why the current starts at zero. As the current grows, \(V_R\) takes a larger share, the back emf falls, and the current grows more and more slowly.

In the example at \(t = 0.5\,\text{s}\): \(V_R = 1.9\times4 = 7.6\,\text{V}\) and \(V_L = 12\times0.37 = 4.4\,\text{V}\). They add to \(12\,\text{V}\), as the loop rule requires.

Where the cell's energy goes

Multiply the loop equation by \(I\). The cell delivers power \(EI\), which splits into \(I^2R\), turned into heat in the resistor, and \(LI\,dI/dt\), stored in the inductor. At first nearly all the power goes into the store. At the end none does, and all of it becomes heat. The store is then full, holding \(\tfrac12 LI_0^2\).

In the example at \(t = 0.5\,\text{s}\), the cell gives \(12\times1.9 = 22.8\,\text{W}\). The resistor turns \(1.9^2\times4 = 14.4\,\text{W}\) into heat. The inductor stores energy at \(I\,V_L = 1.9\times4.4 = 8.4\,\text{W}\). When the current has settled at \(3\,\text{A}\), the inductor holds \(\tfrac12\times2\times3^2 = 9\,\text{J}\).

Two quick rules

Two rules settle most circuits with an inductor at their start and at their end. Just after a switch closes, an inductor that carried no current still carries none, so treat it as an open switch. Long after, the current is steady and the back emf is zero, so treat the inductor as a plain wire. Only the resistance of its own wire remains.

Use the rules on Fig 5.1. Just after K closes, the inductor is an open switch, so the cell sees \(R_1 + R_2 = 8\,\Omega\). It gives \(12/8 = 1.5\,\text{A}\), and all of it flows through \(R_2\). Long after, the inductor is a plain wire across \(R_2\), which then carries nothing. The cell sees only \(R_1\) and gives \(12/4 = 3\,\text{A}\), all through the inductor. In between, the current moves steadily from \(R_2\) into the inductor.

The RC circuit of the chapter Circuits: Kirchhoff's Laws, Networks and RC Transients is the twin of the LR circuit. There the charge on the capacitor grows as \(q_0(1 - e^{-t/RC})\), just as the current here grows as \(I_0(1 - e^{-t/\tau})\). The roles are swapped at the start and at the end, as the table shows.