To measure the current and voltage across a coil, we can use a simple circuit with a resistor and a switch. After closing the switch and applying a voltage source, an induced voltage appears that opposes the applied voltage and delays the rise of the current. This phenomenon is called self-induction.
From Kirchhoff’s loop rule, the applied voltage is equal to the sum of the voltages across the resistor $R_e$ and the inductor $L$:
\[
U_R(t) + U_L(t) = U_0
\]
Substituting Ohm’s law and the induction law gives:
\[
I(t)R_e - L\dot{I}(t) = U_0
\]
The solution for the current is an exponential rise, analogous to charging a capacitor:
\[
\boxed{I(t) = I_0 \left(1 - e^{-\frac{R_e}{L}t}\right)}
\]
From this, we define the time constant $\tau$ as:
\[
\boxed{\tau = \frac{L}{R_e}}
\]
The voltage across the coil is identical to the induced voltage:
\[
U_\mathrm{ind} = -L\dot{I}(t)
\]
Differentiating the current function yields:
\[
\boxed{U(t) = U_0 e^{-\frac{R_e}{L}t}}
\]
Thus, while the current rises exponentially, the voltage across the coil decays exponentially.
When the switch is opened and the source disconnected, the magnetic field collapses, inducing a voltage that maintains the current flow.
In this case, the coil itself acts as a voltage source, with polarity now aligned with the original source.
The loop rule becomes:
\[
U_R(t) + U_L(t) = 0
\]
Substituting Ohm’s law and the induction law gives:
\[
I(t)R_a - L\dot{I}(t) = 0
\]
The solution is again exponential, analogous to discharging a capacitor. For the current:
\[
\boxed{I(t) = I_0 e^{-\frac{R_a}{L}t}}
\]
and for the voltage:
\[
\boxed{U(t) = U_0 e^{-\frac{R_a}{L}t}}
\]
The similarities with capacitor charging and discharging are evident: in both cases, current and voltage follow exponential laws with a characteristic time constant.
Example problem: A coil with inductance $L = 2\,\text{H}$ is connected in series with a resistor of $12\,\Omega$ to a voltage source. How long does it take until the current reaches 70% of its maximum value? Result: 0.2 s