Graded RC and RL transient labs: read a time constant off a charge curve, design a delay, time a discharge, an RL rise and a low-pass cutoff.
On the syllabus
Charging and discharging a capacitor: v = V(1 − e^(−t/RC))
The time constant τ = RC and τ = L/R
Current rise in an inductor
First-order low-pass filter and its cutoff frequency
You will be able to
Read a time constant off a measured curve.
Design an RC delay for a given threshold and time.
Predict the current rise in an RL circuit and its final value.
Design a first-order low-pass filter for a given cutoff.
The first challenge is free. The rest, the viva and the certificate are part of Pro, and everything is here to read.
Graded challenges
Values like 4.7k, 100n or 2.2u are fine. Readings are the number only, in the unit shown. Every check measures the simulation, not a bench: ideal parts within their models, no noise.
1. Read the time constant off a charge curve
A capacitor charging through a resistor rises fast at first, then slower and slower. After one time constant it has covered 63.2% of the way to the supply.
Formulav = V (1 − e^(−t/RC)), τ = RC
Worked example47 kΩ and 1 µF give τ = 47 ms. From a 9 V step the capacitor reaches 63.2% of 9 V, 5.69 V, at 47 ms.
On the scope or meterOn the scope, a curve that rises steeply and then flattens out toward the supply. Read the time where it crosses 63.2%.
Your task
A 5 V step charges C1 (10 µF) through R1 (10 kΩ), starting from 0 V when the simulation starts.
Put the oscilloscope on the capacitor. Measure how long it takes to reach 63.2% of 5 V (3.16 V), in milliseconds.
2. Design an RC delay
A delay circuit waits until the capacitor crosses a threshold. The time to any level follows from the charge curve; half the supply takes ln 2, about 0.693, time constants.
Formulat = RC ln 2, so R = t / (C ln 2)
Worked exampleA 100 ms delay with 10 µF: R = 0.1 / (10 × 10⁻⁶ × 0.693) = 14.4 kΩ. The E12 value 15 kΩ crosses half the supply at 104 ms.
On the scope or meterThe scope crossing of 2.5 V moves later as R grows and earlier as R shrinks.
Your task
C1 is 47 µF. A 5 V step charges it through R1, and a logic input watching the capacitor switches at 2.5 V.
Choose R1 so the capacitor reaches 2.5 V 50 ms after the step, within 5%.
Half the supply takes t = RC × ln 2.
3. Time a discharge
A discharging capacitor falls the same way a charging one rises, toward zero instead of toward the supply. After one time constant 36.8% is left.
Formulav = V e^(−t/RC)
Worked example9 V through 100 kΩ into 2.2 µF: τ = 220 ms, so after 220 ms 3.31 V is left, and after five τ almost nothing.
On the scope or meterThe input square wave falls at 100 ms, and the capacitor voltage decays from there. Time it from the falling edge.
Your task
A 5 Hz square wave between 0 V and 5 V drives R1 (10 kΩ) into C1 (1 µF). It is high for the first 100 ms and low for the next 100 ms.
At 100 ms the input falls and the capacitor starts to discharge from 5 V. Measure how long after that it takes to fall to 36.8% (1.84 V).
4. Current rising in an inductor
An inductor resists a change of current. After a step the current rises along the same kind of curve as a charging capacitor, to a final value set by the resistance alone.
Formulai = (V/R)(1 − e^(−t/τ)), τ = L / R
Worked example12 V, 200 mH and 40 Ω: the current settles at 12 / 40 = 300 mA, with τ = 0.2 / 40 = 5 ms.
On the scope or meterThe voltage across R1 follows the current, so on the scope it rises like an RC charge curve.
Your task
A 5 V step drives L1 (100 mH) in series with R1 (100 Ω).
Measure the final current, once it has settled, and the time for the voltage across R1 to reach 63.2% of its final 5 V.
5. Design a low-pass filter for 1 kHz
An RC low-pass filter passes slow signals and shrinks fast ones. At the cutoff the capacitor’s reactance equals R, and the output is 1/√2 of the input, 3 dB down.
Formulaf = 1 / (2πRC), so C = 1 / (2πRf)
Worked exampleA 500 Hz cutoff with 10 kΩ: C = 1 / (2π × 10 000 × 500) = 31.8 nF. The E12 value 33 nF puts the cutoff at 482 Hz.
On the scope or meterAt the cutoff, the output sine is smaller than the input and lags it by 45°.
Your task
A 1 kHz sine of 1 V peak drives R1 (1.6 kΩ) into C1. The output is across C1.
Choose C1 so that 1 kHz is the cutoff: the output amplitude is 1/√2 of the input, 0.707 V peak, within 5%.
The cutoff of a first-order RC filter is f = 1 / (2πRC).
Viva questions
The questions an examiner asks at the bench. 75% passes; every answer comes with why.