Question 5
5
The variation with potential difference V of the charge Q on one of the plates of a capacitor is shown in Fig. 5.1. The capacitor is connected to an 8.0 V power supply and two resistors R and S as shown in Fig. 5.2. The resistance of R is \(25 \mathrm{k} \Omega\) and the resistance of S is \(220 \mathrm{k} \Omega\). The switch can be in either position X or position Y .
structured5 marks
Question 5(b)
5(b)
The switch is now moved to position Y.
structured5 marks
Question 5(b)(i)
5(b)(i)
Show that the time constant of the discharge circuit is 3.3 s .
Mediumstructured2 marks
Answer
(t=) R C C1 ( t= ) \(220 \times 10^{3} \times\left(1.2 \times 10^{-4} / 8.0\right)=3.3 \mathrm{~s}\) A1
Question 5(b)(ii)
5(b)(ii)
The fully charged capacitor in (a) stores energy E. Determine the time t taken for the stored energy to decrease from E to E / 9.
Hardstructured3 marks
Answer
\(E \propto V^{2}\) C1 (so time to) \(V_{\mathrm{o}} / 3\) \[ V=V_{0} e^{-t / R C} \] C1 \[ \begin{aligned} \frac{V_{o}}{3}=V_{o} e^{-t / 3.3} \frac{1}{3}=e^{-t / 3.3} \end{aligned} \] C1 \(\mathrm{t}=3.6 \mathrm{~s}\) A1
Question 5
5
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. 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 .
structured3 marks
Question 5(a)
5(a)
3 marks
Question 5(a)(iii)
5(a)(iii)
Calculate the time constant \(\tau\) of the discharge circuit.
Easystructured1 marks
Answer
\(\tau=R C\) C1 \[ \begin{aligned} =5600 \times 470 \times 10^{-6} =2.6 \mathrm{~s} \end{aligned} \] A1
Question 5(a)(iv)
5(a)(iv)
On Fig. 5.2, sketch a line to show the variation with t of the current I in wire P as the capacitor discharges.
Mediumstructured2 marks
Answer
line with negative gradient throughout passing through ( \(0, I_{0}\) ) B1 exponential decay curve asymptotic to t-axis B1
Question 5
5
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. 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.
structured5 marks
Question 5(a)
5(a)
Show that the capacitance of the capacitor is \(1.5 \mu \mathrm{~F}\).
Mediumstructured3 marks
Answer
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
Question 5(b)
5(b)
Determine the resistance of R.
Mediumstructured0 marks
Answer
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)
Question 5(d)
5(d)
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 \(t=5.0 \mathrm{~s}\).
Mediumstructured2 marks
Answer
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
Question 5
5
3 marks
Question 5(c)
5(c)
Two capacitors of capacitances \(22 \mu \mathrm{~F}\) and \(47 \mu \mathrm{~F}\), and a resistor of resistance \(2.7 \mathrm{M} \Omega\), are connected into the circuit of Fig. 5.2. The battery has an e.m.f. of 12 V .
structured3 marks
Question 5(c)(iii)
5(c)(iii)
The two-way switch is now moved to position Y . Determine the time taken for the potential difference (p.d.) across the \(22 \mu \mathrm{~F}\) capacitor to become 6.0 V .
Hardstructured3 marks
Answer
initial p.d. (across \(22 \mu \mathrm{~F})=12 \times(15 / 22)\) = 8.2 V or final p.d. across both capacitors \(=6.0 \times(22 / 15)\) = 8.8 V C1 \(V=V_{0} \exp \left[-t /\left(2.7 \times 10^{6} \times 15 \times 10^{-6}\right)\right]\) C1 \[ \begin{aligned} & 6.0=8.2 \exp \left[-t /\left(2.7 \times 10^{6} \times 15 \times 10^{-6}\right)\right] & \text { or } \end{aligned} \] \[ 8.8=12 \exp \left[-t /\left(2.7 \times 10^{6} \times 15 \times 10^{-6}\right)\right] \] \(t=13 \mathrm{~s}\) A1
Question 6
6
A capacitor of capacitance C and a resistor of resistance R are connected as shown in Fig. 6.1. Initially, the capacitor is charged and the switch is open. The switch is closed at time t=0. Fig. 6.2 and Fig. 6.3 show, respectively, the variations with t of the charge Q on the capacitor and the potential difference (p.d.) V across the resistor.
structured6 marks
Question 6(a)
6(a)
Explain the shape of the line in Fig. 6.3 representing the variation of V with t.
Mediumstructured3 marks
Answer
p.d. across resistor = p.d. across capacitor - current (in resistor) proportional to p.d. across it - current causes capacitor to lose charge - charge (on capacitor) proportional to p.d. so p.d. decreases Any two points, 1 mark each B2 rate of change of p.d. decreases as p.d. decreases B1
Question 6(b)
6(b)
Use Fig. 6.2 to show that the time constant of the circuit in Fig. 6.1 is 5.5 s .
Mediumstructured3 marks
Answer
\(Q_{0}=0.90 \mathrm{mC}\) and at t= one time constant, \(Q=Q_{0} \exp (-1)\) B1 at t= one time constant, \(Q=0.90 \exp (-1)=0.33 \mathrm{mC}\) M1 evidence of graph reading: when \(Q=0.33 \mathrm{mC}, t=5.5 \mathrm{~s}\) A1 or evidence of two correct sets of readings for Q and t from the graph (B1) correct substitution of Q and t values into \(Q_{2}=Q_{1} \exp \left[\left(t_{1}-t_{2}\right) / \tau\right]\) (M1) calculation to give \(\tau=5.5 \mathrm{~s}\) (A1) or read-off of half-life as 3.75 s (B1) use of \(Q=Q_{0} \exp (-t / \tau)\) to show that \(\tau=\) half-life \(/ \ln 2\) (M1) \(\tau=3.75 / \ln 2=5.4 \mathrm{~s}\) (A1) 6(b) or tangent drawn on Q-t graph and value of Q at exact same time as tangent read from graph (M1) gradient of tangent correctly calculated (A1) \(\tau=Q /\) gradient used to correctly calculate a value for \(\tau\) as 5.5 s (A1)
Question 5
5
Part of an electric circuit is shown in Fig. 5.1. The circuit is used to produce half-wave rectification of an alternating voltage of potential difference (p.d.) \(V_{\text {IN }}\). The output p.d. across the \(14 \mathrm{k} \Omega\) resistor is \(V_{\text {OUT }}\).
structured4 marks
Question 5(b)
5(b)
Fig. 5.2 shows the variation with time t of \(V_{\text {IN }}\). Fig. 5.3 shows the variation with t of \(V_{\text {OUT }}\).
structured4 marks
Question 5(b)(ii)
5(b)(ii)
Show that the time constant \(\tau\) for the discharge of the capacitor through the resistor is 0.038 s .
Mediumstructured2 marks
Answer
\[ \begin{aligned} V=V_{0} \exp (-t / R C) \text { and } \tau=R C \text { or } V=V_{0} \exp (-t / \tau) \end{aligned} \] C1 \(3.25=5.50 \exp (-0.020 / \tau)\) leading to \(\tau=0.038 \mathrm{~s}\) A1
Question 5(b)(iii)
5(b)(iii)
Calculate the capacitance of C. Give a unit with your answer. capacitance = unit
Mediumstructured2 marks
Answer
\(\tau=R C\) C1 capacitance =0.038 / 14000 \[ =2.7 \times 10^{-6} \mathrm{~F} \] A1
Question 6
6
A capacitor C is charged so that the potential difference (p.d.) V across its terminals is 8.0 V . The capacitor is connected into the circuit of Fig. 6.1. The switch is initially open. The switch is closed at time t=0.
structured3 marks
Question 6(b)
6(b)
Fig. 6.3 shows the variation with t of \(-\ln \left(\frac{V}{8.0 \mathrm{~V}}\right)\).
structured3 marks
Question 6(b)(i)
6(b)(i)
Show that, when t is equal to one time constant, the value of \(-\ln \left(\frac{V}{8.0 \mathrm{~V}}\right)\) is equal to 1.0 .
Mediumstructured2 marks
Answer
\(V=V_{0} \exp (-t / R C)\) and \(\tau=R C\) C1 \[ \begin{aligned} V=V_{0} \exp (-t / \tau) V_{0}=8.0 \mathrm{~V} \text {, and at one time constant, } t=\tau V / 8.0=\exp (-\tau / \tau), \text { so } \ln (V / 8.0)=-1.0 \text { or }-\ln (V / 8.0)=1.0 \end{aligned} \] A1
Question 6(b)(ii)
6(b)(ii)
Determine the time constant \(\tau\) of the circuit in Fig. 6.1.
Mediumstructured1 marks
Answer
[ t read from graph at \(-\ln (V / 8.0)=1.0]: \tau=3.2 \mathrm{~s}\) A1
Question 6(b)(iii)
6(b)(iii)
Calculate the resistance of resistor R. resistance = \(\Omega\)
Mediumstructured0 marks
Answer
\(\tau=R C\) C1 \[ \begin{aligned} R =3.2 /\left(5.6 \times 10^{-5}\right) =5.7 \times 10^{4} \Omega \end{aligned} \] A1
Question 6
6
Fig. 6.1 shows a capacitor of capacitance C connected in series with a resistor of resistance R. Initially the switch is open and there is a p.d. of 12 V across the capacitor. At time t=0, the switch is closed so that there is a current I in the resistor. Fig. 6.2 shows the variation of I with t.
structured10 marks
Question 6(a)
6(a)
Explain the shape of the line in Fig. 6.2.
Mediumstructured3 marks
Answer
p.d. across capacitor proportional to charge on capacitor - p.d. across capacitor = p.d. across resistor - current in resistor proportional to p.d. across resistor - current in resistor = rate of decrease of charge on capacitor Any two points, 1 mark each B2 charge proportional to current so rate of decrease of current decreases as current decreases (therefore exponential shape) B1
Question 6(b)
6(b)
Use Fig. 6.2 to determine:
structured5 marks
Question 6(b)(i)
6(b)(i)
resistance R
Mediumstructured2 marks
Answer
\[ \begin{aligned} R =V / I =12 /\left(0.13 \times 10^{-3}\right) \end{aligned} \] C1 \(=9.2 \times 10^{4} \Omega\) A1
Question 6(b)(ii)
6(b)(ii)
the time constant \(\tau\) of the circuit in Fig. 6.1.
Mediumstructured3 marks
Answer
correct read-off of at least one pair of values for I and t C1 attempted read-off of t when \(I=0.048 \mathrm{~mA}\) or substitution of a correct pair of values of I and t into \(I=0.13 \exp (-t / \tau)\) C1 \(\tau=4.3 \mathrm{~s}\) A1
Question 6(c)
6(c)
Use your answers in (b) to determine capacitance C.
Hardstructured2 marks
Answer
\(\tau=R C\) C1 \[ \begin{aligned} C =\tau / R=4.3 /\left(9.2 \times 10^{4}\right) =4.7 \times 10^{-5} \mathrm{~F} \end{aligned} \] A1