RL circuit on the oscilloscope
Curricula: APC, Sek II
Aim¶
Observe the rise and decay of the current in an RL circuit and measure the time constant \(L / R\).
The bench¶
AC voltage source — a square wave from 0 to 5 V (amplitude 2.5 V, DC offset 2.5 V), 10 Hz. A 1 H inductor (winding 10 Ω) and a 100 Ω resistor in series. Oscilloscope: CH1 — the source, CH2 — resistor (the voltage across it is proportional to the current), DUAL mode, 2 V/div, 5 ms/div, triggered on CH1 at 2.5 V.
Procedure¶
- Examine the CH2 waveform.
- Measure the time it takes the voltage across the resistor to reach 63% of its steady-state value.
- In the resistor's settings, change 100 Ω to 220 Ω and repeat.
Expected result
The current rises and decays exponentially. The steady-state voltage across the resistor is about 4.5 V (5 V · 100 / 110); 63% of it — about 2.9 V — is reached after \(\tau = L / R\) ≈ 9 ms, a little under two divisions. The half-period of 50 ms is more than five \(\tau\), so the current has time to settle; at a higher frequency it would not reach 4.5 V, and the 63% would have to be taken of a smaller value. With 220 Ω, \(\tau\) falls to about 4.4 ms.
Questions¶
- Why does the time constant decrease as the resistance increases?
- How is an RL circuit similar to an RC circuit, and how does it differ?
Answers
- At the first moment the whole source voltage is across the inductor, so the current starts rising at the rate \(U / L\) whatever the resistance. A larger resistance lowers the steady current \(U / R\), and at the same starting rate the current gets close to it sooner, hence \(\tau = L / R\). With 220 Ω the total resistance is about 230 Ω instead of 110 Ω, and \(\tau\) falls to about 4.4 ms.
- In both circuits the voltages and the current change exponentially with a time constant: 63% of the change after one \(\tau\), practically complete after five; the resistor waveform here has the same shape as the capacitor waveform in an RC circuit. They differ in what cannot jump and in the steady state: in RC the capacitor voltage is continuous and the current dies away to zero, in RL the inductor current is continuous and settles at \(U / R\). \(\tau = R \cdot C\) grows with the resistance, \(\tau = L / R\) falls.