Question

A full-wave rectifier has a load resistor of \(1\text{ k}\Omega\) and a capacitor of \(100\ \mu\text{F}\). Doubling the load resistor will _ _ _ _ .

• A. double ripple voltage
• B. halve ripple voltage
• C. no change
• D. quadruple ripple voltage

Hint:

The ripple voltage in a capacitor filter is inversely proportional to the load resistance (\(V_r \propto \frac{1}{R_L}\)). Doubling the load resistance reduces the current draw, which halves the ripple voltage.

Answer 1

The Correct Option is B

Understanding the Concept: A full-wave rectifier converts alternating current (AC) into pulsating direct current (DC). To smooth out these voltage variations and deliver a steady DC output, a filter capacitor (\(C\)) is connected in parallel with the load resistor (\(R_L\)). The residual periodic variation remaining on the DC output voltage is called the ripple voltage (\(V_r\)). For a full-wave rectifier equipped with a parallel capacitor filter, the peak-to-peak ripple voltage can be calculated using the standard approximation formula: \[ V_r = \frac{I_{DC}}{2 f C} = \frac{V_{DC}}{2 f R_L C} \] Where:
• \(V_{DC}\) represents the average output DC voltage level.
• \(f\) represents the frequency of the input AC power supply.
• \(R_L\) represents the resistance value of the connected load resistor.
• \(C\) represents the capacitance value of the filter capacitor.

Step 1: Analyzing the mathematical proportionality between ripple voltage and load resistance.
From the ripple voltage formula, we can see that if the average output voltage, input frequency, and filter capacitance are held constant, the peak-to-peak ripple voltage is inversely proportional to the value of the load resistor: \[ V_r \propto \frac{1}{R_L} \]

Step 2: Calculating the effect of doubling the load resistance.
Let the initial ripple voltage be \(V_{r1} = \frac{k}{R_{L1}}\). If the load resistor is doubled, the new resistance value becomes \(R_{L2} = 2 R_{L1}\). Substituting this new value into our proportionality relation gives: \[ V_{r2} = \frac{k}{2 R_{L1}} = \frac{1}{2} \cdot \left( \frac{k}{R_{L1}} \right) = \frac{1}{2} \cdot V_{r1} \] This shows that doubling the load resistance cuts the ripple voltage exactly in half. This matches Option (B).