AC Voltage Applied to a Capacitor

The effective resistance offered by a capacitor to the alternating current is called capacitive reactance.

X_C​ = \[\frac {1}{2πfC}\]​ (∝ 1/f)

A capacitor of capacitance C is connected to an AC source.

The applied voltage at any instant is: v = v_m ​sin ωt

Where:

• v_m​ = peak voltage (V)
• ω = angular frequency (rad/s)
• t = time (s)

Step 1: Voltage across the capacitor equals applied voltage:

Step 2: Express charge:

Step 3: Find instantaneous current (i = \[\frac {dq}{dt}\]):

Step 4: Rewrite using cos⁡ ωt = sin⁡ ⁣(ωt + \[\frac {π}{2}\]):

Where:

Instantaneous power:

Average over one complete cycle:

Question

A lamp is connected in series with a capacitor. Predict your observations for DC and AC connections. What happens in each case if the capacitance of the capacitor is reduced?

Solution

Case 1 — DC Source

• When a DC source is connected to a capacitor, the capacitor gets charged.
• After charging, no current flows in the circuit.
• Therefore, the lamp will not glow.
• There will be no change even if C is reduced — the lamp still does not glow.

Case 2 — AC Source

• With an AC source, the capacitor offers capacitive reactance (\[\frac {1}{ωC}\]).
• Current flows in the circuit.
• Consequently, the lamp will shine.
• Reducing C will increase the reactance (X_C = \[\frac {1}{ωC}\]).
• Therefore, the lamp will shine less brightly than before.

Question

A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current?

Solution

Step 1: Capacitive Reactance

Step 2: RMS Current

Step 3: Peak Current

Step 4: Effect of Doubling Frequency

• If the frequency is doubled, the capacitive reactance is halved.
• Consequently, the current is doubled.

Question

A light bulb and an open coil inductor are connected to an AC source through a key. The switch is closed, and after some time, an iron rod is inserted into the interior of the inductor. The glow of the light bulb:

1. increases
2. decreases
3. is unchanged

Give your answer with reasons.

Solution

When the iron rod is inserted:

1. The magnetic field inside the coil magnetises the iron, increasing the magnetic field inside it.
2. Hence, the inductance of the coil increases.
3. Consequently, the inductive reactance of the coil increases.
4. As a result, a larger fraction of the applied AC voltage appears across the inductor, leaving less voltage across the bulb.
5. Therefore, the glow of the light bulb decreases.