Physics 9702 · Paper 4

ln Graphs & Smoothing

Two standard packagings of the discharge equation: take logarithms and the curve becomes a straight line; put the capacitor after a rectifier and the decay becomes a design feature.

Straightening the curve with ln

Take ln of $Q = Q_0 e^{-t/RC}$ : $\ln Q = \ln Q_0 - \dfrac{t}{RC}$ — a straight line with intercept $\ln Q_0$ and gradient $-1/RC$ .
A 2023 question ran the whole chain backwards from such a graph: intercept 2.9 → $Q_0 = e^{2.9} = 18.2\ \mu\text{C}$ → $C = Q_0/12\ \text{V} = 1.5\ \mu\text{F}$ ; then gradient $-0.25 = -1/RC$ → $R = 2.7\times10^{6}\ \Omega$ (9702/42/F/M/23 Q5).
Its final part was the best test of understanding: add a second identical resistor in parallel. Resistance halves, so the gradient doubles — but the intercept is the starting charge, which does not move (9702/42/F/M/23 Q5(d)). A variant used $-\ln(V/V_0)$ on the axis, where the value 1.0 is reached exactly at $t = \tau$ (9702/42/O/N/23 Q6(b)).

A rectifier turns a.c. into bumpy one-way voltage. A large capacitor across the output charges at each peak and then feeds the load between peaks — the output only sags a little before the next peak arrives. Asked what the capacitor does, the one-word answer “smoothing” scores, and it has been asked four times in three years (9702/41/M/J/23 Q5(a)(ii)) (9702/41/O/N/24 Q6(b)(ii)) (9702/41/M/J/25 Q6(a)(ii)).
The sag between peaks is a discharge curve, so every smoothing question becomes last lesson's algebra: read the ripple (say 12 V falling to 8.0 V in 7.3 ms), solve $8.0 = 12\,e^{-t/\tau}$ for $\tau = 0.018$ s, then split $\tau = RC$ for whichever of $R$ or $C$ is unknown (9702/41/M/J/25 Q6(b)(iv)) (9702/41/O/N/24 Q6(c)).
Design intuition: bigger $RC$ compared with the time between peaks → smaller ripple. (The rectifier circuits themselves — diodes, bridges, full-wave — belong to the Alternating Currents chapter.)

$\ln Q$ – $t$ : intercept $\ln Q_0$ , gradient $-1/RC$ . Smoothing = discharge between peaks; the ripple is data for $\tau$ .

ln Graphs & Smoothing

Two standard packagings of the discharge equation: take logarithms and the curve becomes a straight line; put the capacitor after a rectifier and the decay becomes a design feature.

Take ln of $Q = Q_0 e^{-t/RC}$ : $\ln Q = \ln Q_0 - \dfrac{t}{RC}$ — a straight line with intercept $\ln Q_0$ and gradient $-1/RC$ .
A 2023 question ran the whole chain backwards from such a graph: intercept 2.9 → $Q_0 = e^{2.9} = 18.2\ \mu\text{C}$ → $C = Q_0/12\ \text{V} = 1.5\ \mu\text{F}$ ; then gradient $-0.25 = -1/RC$ → $R = 2.7\times10^{6}\ \Omega$ (9702/42/F/M/23 Q5).
Its final part was the best test of understanding: add a second identical resistor in parallel. Resistance halves, so the gradient doubles — but the intercept is the starting charge, which does not move (9702/42/F/M/23 Q5(d)). A variant used $-\ln(V/V_0)$ on the axis, where the value 1.0 is reached exactly at $t = \tau$ (9702/42/O/N/23 Q6(b)).

A rectifier turns a.c. into bumpy one-way voltage. A large capacitor across the output charges at each peak and then feeds the load between peaks — the output only sags a little before the next peak arrives. Asked what the capacitor does, the one-word answer “smoothing” scores, and it has been asked four times in three years (9702/41/M/J/23 Q5(a)(ii)) (9702/41/O/N/24 Q6(b)(ii)) (9702/41/M/J/25 Q6(a)(ii)).
The sag between peaks is a discharge curve, so every smoothing question becomes last lesson's algebra: read the ripple (say 12 V falling to 8.0 V in 7.3 ms), solve $8.0 = 12\,e^{-t/\tau}$ for $\tau = 0.018$ s, then split $\tau = RC$ for whichever of $R$ or $C$ is unknown (9702/41/M/J/25 Q6(b)(iv)) (9702/41/O/N/24 Q6(c)).
Design intuition: bigger $RC$ compared with the time between peaks → smaller ripple. (The rectifier circuits themselves — diodes, bridges, full-wave — belong to the Alternating Currents chapter.)

$\ln Q$ – $t$ : intercept $\ln Q_0$ , gradient $-1/RC$ . Smoothing = discharge between peaks; the ripple is data for $\tau$ .