Physics 9702 · Paper 4

Smoothing

Rectified voltage is one-way but bumpy. A capacitor across the load flattens the bumps — and turns every exam question into a discharge-curve calculation you already know.

The circuit and the ripple

A capacitor across the load holds the voltage up between peaks: it charges at each hump and discharges gently through R until the next one. The dips that remain are the ripple — smaller when the time constant RC is large compared with the time between peaks.

The capacitor sits in parallel with the load — drawing it there was a one-marker (9702/42/F/M/24 Q4(c)(i)). It charges near each peak, then feeds the load while the rectified voltage dips, discharging through $R$ until the next peak lifts it again. Asked its purpose, the one-word answer “smoothing” scores — four sessions running (9702/41/M/J/23 Q5(a)(ii)) (9702/41/O/N/24 Q6(b)(ii)) (9702/41/M/J/25 Q6(a)(ii)).
What remains of the bumps is the ripple. The sag between peaks is exactly the exponential discharge from the Capacitance chapter, so bigger $RC$ compared with the time between peaks means smaller ripple — and every smoothing calculation is $V = V_0 e^{-t/RC}$ applied to the sag.
One rubric detail with its own mark: read the two ( $t$ , $V$ ) points from within the same discharge cycle — for example (5.0 ms, 4.0 V) and (13.0 ms, 3.2 V) gave $\tau = 36$ ms (9702/42/F/M/24 Q4(c)(ii)). Points from different sags mix two different curves.
The full chains — ripple 12→8.0 V in 7.3 ms giving $\tau = 0.018$ s and then $R = \tau/C = 32\ \Omega$ — were worked in the Capacitance chapter's smoothing lesson; the physics here is identical (9702/41/M/J/25 Q6(b)(iv)).

Your turn— tap to reveal the worked answer (9702/41/M/J/23 Q5(c))

A full-wave rectifier with a smoothing capacitor is fed with a square-wave input of constant magnitude 12 V. What does the output look like?

The input's magnitude never changes — the negative halves are flipped up to the same 12 V — so the rectified output is already flat: $V_{OUT}$ is constant, and the capacitor has nothing left to smooth.

Capacitor in parallel with the load; “smoothing” scores; ripple = discharge curve; same-cycle read-offs; bigger $RC$ (relative to the gap between peaks) = flatter output.

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Smoothing

Rectified voltage is one-way but bumpy. A capacitor across the load flattens the bumps — and turns every exam question into a discharge-curve calculation you already know.

The circuit and the ripple

The capacitor sits in parallel with the load — drawing it there was a one-marker (9702/42/F/M/24 Q4(c)(i)). It charges near each peak, then feeds the load while the rectified voltage dips, discharging through $R$ until the next peak lifts it again. Asked its purpose, the one-word answer “smoothing” scores — four sessions running (9702/41/M/J/23 Q5(a)(ii)) (9702/41/O/N/24 Q6(b)(ii)) (9702/41/M/J/25 Q6(a)(ii)).
What remains of the bumps is the ripple. The sag between peaks is exactly the exponential discharge from the Capacitance chapter, so bigger $RC$ compared with the time between peaks means smaller ripple — and every smoothing calculation is $V = V_0 e^{-t/RC}$ applied to the sag.
One rubric detail with its own mark: read the two ( $t$ , $V$ ) points from within the same discharge cycle — for example (5.0 ms, 4.0 V) and (13.0 ms, 3.2 V) gave $\tau = 36$ ms (9702/42/F/M/24 Q4(c)(ii)). Points from different sags mix two different curves.
The full chains — ripple 12→8.0 V in 7.3 ms giving $\tau = 0.018$ s and then $R = \tau/C = 32\ \Omega$ — were worked in the Capacitance chapter's smoothing lesson; the physics here is identical (9702/41/M/J/25 Q6(b)(iv)).

Your turn— tap to reveal the worked answer (9702/41/M/J/23 Q5(c))

A full-wave rectifier with a smoothing capacitor is fed with a square-wave input of constant magnitude 12 V. What does the output look like?

Capacitor in parallel with the load; “smoothing” scores; ripple = discharge curve; same-cycle read-offs; bigger $RC$ (relative to the gap between peaks) = flatter output.