CAIE A-Level Physics 19.3 Capacitor Discharge

Practise analysing capacitor-discharge graphs, calculating the time constant tau = RC and applying exponential equations to determine circuit quantities.

Exam points

analyse capacitor-discharge graphs for potential difference, charge and current
calculate and interpret τ=RC
apply exponential discharge equations to determine time, initial/final quantities or circuit parameters

Covered syllabus points

19.3.1—The variation with time of potential difference, charge and current 19.3.2—Τ = RC for the time constant for a capacitor discharging through a resistor 19.3.3—Equations of the form x = x0 e–(t / RC) where x could represent current

[Maximum number: 4]

A capacitor of capacitance $470 \mu \mathrm{~F}$ is connected to a battery of electromotive force (e.m.f.) 24 V in the circuit of Fig. 5.1.

Fig. 5.1

The two-way switch S is initially at position X.
P and Q are identical long straight wires, each with a resistance of $5.6 \mathrm{k} \Omega$ . These wires are placed near to, and parallel to, each other. Wire Q is connected to a voltmeter.

At time t=0, switch S is moved to position Y so that the capacitor discharges through wire P .

(a)

Calculate the current $I_{0}$ in wire P at time t=0.

I_{0}=

[ 1 ]

$\begin{aligned} I_{0} =24 / 5600 =4.3 \times 10^{-3} \mathrm{~A} \end{aligned}$

A1

(b)

Calculate the time constant $\tau$ of the discharge circuit.

\tau=

[ 1 ]

$\tau=R C$

C1

$\begin{aligned} =5600 \times 470 \times 10^{-6} =2.6 \mathrm{~s} \end{aligned}$

A1

(c)

On Fig. 5.2, sketch a line to show the variation with t of the current I in wire P as the capacitor discharges.

Fig. 5.2

[ 2 ]

line with negative gradient throughout passing through ( $0, I_{0}$ )

B1

exponential decay curve asymptotic to t-axis

B1

[Maximum number: 8]

A capacitor, a battery of electromotive force (e.m.f.) 12 V , a resistor R and a two-way switch are connected in the circuit shown in Fig. 5.1.

Fig. 5.1

The switch is initially in position S . When the capacitor is fully charged, the switch is moved to position T so that the capacitor discharges. At time t after the switch is moved the charge on the capacitor is Q.

The variation with t of $\ln (Q / \mu C)$ is shown in Fig. 5.2.

Fig. 5.2

(a)

Show that the capacitance of the capacitor is $1.5 \mu \mathrm{~F}$ .

[ 3 ]

from graph $\ln Q=2.9$ $\text { (so } Q=18.2 \mu \mathrm{C} \text { ) }$

B1

C=Q / V

C1

$=18.2 / 12=1.5 \mu \mathrm{~F}$

A1

(b)

Determine the resistance of R.

resistance =

[ 3 ]

gradient =-0.25

C1

gradient =-1 / R C

C1

$\begin{aligned} R =1 /\left(0.25 \times 1.5 \times 10^{-6}\right) =2.7 \times 10^{6} \Omega \end{aligned}$

A1

or $\frac{Q}{Q_{0}}=e^{-t / C R}$ or $\ln Q-\ln Q_{0}=\frac{-t}{C R}$

(C1)

e.g. $\frac{4.95}{18.2}=e^{-5.2 /\left(1.5 \times 10^{-6} R\right)}$ or $1.6-2.9=5.2 /\left(1.5 \times 10^{-6} R\right)$

(C1)

$R=2.7 \times 10^{6} \Omega$

(A1)

(c)

A second identical resistor is now connected in parallel with R.

The switch is initially in position S . When the capacitor is fully charged, the switch is moved to position T so that the capacitor discharges. At time t after the switch is moved the charge on the capacitor is Q.

On Fig. 5.2, sketch a line to show the variation of $\ln (Q / \mu \mathrm{C})$ with t between time t=0 and time $t=5.0 \mathrm{~s}$ .

[ 2 ]

straight line with different negative gradient starting from ( 0,2.9 )

M1

straight line between t=0 and at least $t=5.0 \mathrm{~s}$ with twice the gradient of the original line

A1

All question bank results loaded

Covered syllabus points

19.3 Discharging a capacitor question 1

Question (a)

Question (b)

Question (c)

19.3 Discharging a capacitor question 2

Question (a)

Question (b)

Question (c)