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Chapter 2: Problem 13

If the lond resistance of a capxitor-filtered full-wave rectifier is redticed, the ripple voltage (a) increases (b) decreases (c) is not affected (d) has a different frequency

Short Answer

Expert verified
The ripple voltage (Vr) of a capacitor-filtered full-wave rectifier is inversely proportional to the load resistance (R), as given by the formula: \(Vr = \frac{V_m}{2\pi fRC}\). When the load resistance (R) is reduced, the ripple voltage (Vr) increases. So, the correct answer is (a) increases.

Step by step solution

01

Understand the components of a capacitor-filtered full-wave rectifier

A capacitor-filtered full-wave rectifier consists of a transformer, a full-wave diode bridge, a capacitor, and a load resistance (R). The purpose of the capacitor is to reduce the ripple voltage across the load resistor.
02

Recall the formula for ripple voltage

The ripple voltage (Vr) in a full-wave rectifier is given by the formula: \[Vr = \frac{V_m}{2\pi fRC}\] where \(V_m\) is the peak voltage, f is the frequency, and C is the capacitance of the capacitor.
03

Analyze the effect of reduced load resistance on ripple voltage

Since we are asked how the ripple voltage changes as load resistance (R) is reduced, let's examine the formula: \[Vr = \frac{V_m}{2\pi fRC}\] Here, we can observe that the ripple voltage (Vr) is inversely proportional to the load resistance (R). So, when the load resistance (R) decreases, the ripple voltage (Vr) increases. Thus, the correct answer is: (a) increases

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