JEE Main 2021 · 31 August · evening · Single correct · +4 / −1
Statement - I :
To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect a capacitor across the output parallel to the load R L .
Statement - II :
To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect an inductor in series with R L .
In the light of the above statements, choose the most appropriate answer from the options given below :
- (a) Statement I is true but Statement II is false
- (b) Statement I is false but Statement II is true
- (c) Both Statement I and Statement II are false
- (d) Both Statement I and Statement II are true
Answer and solution
Answer (d) Both Statement I and Statement II are true
Solution
Physics sir · 4 steps
A filter must pass dc and block ac: a capacitor does it in parallel, an inductor in series.
- What needs fixing. The output of a full wave rectifier is a steady dc part plus an ac ripple. A filter must send the dc part through R_L and keep the ripple out of it.
- Capacitor in parallel. A capacitor has reactance X_C = 1/(ω C), which is very small for the ripple and infinite for dc. In parallel with R_L it gives the ripple an easy path around the load. It also charges at each peak and discharges slowly through R_L in between, so the voltage stays nearly constant. Statement I is true.
- Inductor in series. An inductor has reactance X_L = ω L, which is zero for dc and large for the ripple. In series with R_L it lets the dc current through and opposes any change in current, so the ripple is smoothed out. Statement II is true.
- So. Both methods smooth the pulsating output into nearly steady dc, so both statements are true. The answer is (d).
The idea to keep. Ask where each component sends the ripple. A capacitor shorts ac, so it goes in parallel with the load. An inductor blocks ac, so it goes in series. The usual practical filter combines both.
Answer from the NTA final answer key. Solution by PrepWiser.