Question 4 of 16intermediate🔧 ApplyNumerical3 marks

A 15.0 μ\muF 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?

Correct Answer

Given: C = 15.0 μF = 15.0×1015.0 \times 10⁻⁶ F, V = 220 V, ν = 50 Hz

Step 1: Capacitive Reactance Xc = 1/(2πνC) = 1/[2π × 50×15.050 \times 15.0 × 10⁻⁶] = 212 Ω

Step 2: RMS Current I = V/Xc = 220/212 = 1.04 A

Step 3: Peak Current im = 2\sqrt{2} × I = 1.41×1.041.41 \times 1.04 = 1.47 A

Step 4: Effect of Doubling Frequency When frequency is doubled, capacitive reactance is halved (since Xc ∝ 1/ν), and consequently, the current is doubled.

Exercise: Example 7.4 | Q: Example 7.4 | (Page 9)
For More Understanding

Explanation

This is a straightforward application of capacitive reactance formula Xc = 1/(2πνC) and Ohm's law for AC circuits. The textbook provides the complete solution showing that Xc = 212 Ω, Irms = 1.04 A, and Ipeak = 1.47 A. The key concept is that capacitive reactance is inversely proportional to frequency, so doubling frequency halves the reactance and doubles the current.

Solution Steps

  1. Step 1: Calculate Xc using Xc = 1/(2πνC)

  2. Step 2: Calculate RMS current using I = V/Xc

  3. Step 3: Calculate peak current using im = 2\sqrt{2} × I

  4. Step 4: Analyze effect of doubling frequency on Xc and I

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