A full-wave rectifier uses a capacitor filter across a load resistance \(R_L = 2.5\ \mathrm{k\Omega}\). The time constant \(CR_L\) of the filter is to be 0.5 s. The capacitance \(C\) needed is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(200\ \mu\mathrm{F}\)
Between two peaks, the filter capacitor discharges through the load. The product \( CR_L \) is the time constant. A larger time constant makes the voltage fall more slowly.
Given:
Time constant, \( \tau=0.5\ \mathrm{s} \)
Load resistance, \( R_L=2.5\ \mathrm{k\Omega}=2500\ \Omega \)
\( \begin{aligned} C &= \frac{\tau}{R_L} \\ &= \frac{0.5}{2500}\ \mathrm{F} \\ &= 2\times10^{-4}\ \mathrm{F} \\ &= 200\ \mu\mathrm{F} \end{aligned} \)
From the above, \( C=200\ \mu\mathrm{F} \). Hence, a capacitor of 200 μF gives the required time constant.
NCERT (2026-27):
Application of Junction Diode as a Rectifier – “… The rate of fall of the voltage across the capacitor depends inversely upon the product of capacitance C and the effective resistance RL used in the circuit and is called the time constant. …”