Question 1
1
Question 1(b)
1(b)
Define potential difference.
Answer
( p.d. \(=) \frac{\text { work (done) or energy (transformed) (from electrical to other forms) }}{\text { charge }}\)
Question bank
A-Level CAIE Physics 9 2 Potential Difference And Power Question Bank
Practice A-Level CAIE Physics 9 2 Potential Difference And Power questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.
Question 1
1
Which physical quantity would result from a calculation in which a potential difference is multiplied by an electric charge? electric current electric energy electric field strength electric power
Answer
B
Question 2
2
What is a possible unit for the product V I, where V is the potential difference across a resistor and I is the current through the same resistor? newton per second ( \(\mathrm{N} \mathrm{s}^{-1}\) ) newton second (Ns) newton metre ( Nm ) newton metre per second ( \(\mathrm{Nm} \mathrm{s}^{-1}\) )
Answer
D
Question 3
3
Which physical quantity would result from a calculation in which a potential difference is multiplied by an electric charge? electric current electric energy electric field strength electric power
Answer
B
Question 3
3
Lightning occurs when charge builds up in the atmosphere, creating a potential difference between the ground and the atmosphere. During a lightning strike there is an average current of \(3.3 \times 10^{4} \mathrm{~A}\) for a time of \(2.6 \times 10^{-5} \mathrm{~s}\).
Question 3(b)
3(b)
The potential difference between the ground and the atmosphere is \(3.0 \times 10^{7} \mathrm{~V}\). Calculate the average power, in GW, transferred during the lightning strike. power = GW
Answer
\[ \begin{aligned} P=I V \text { or } P=V Q / t \text { or } V=I R \text { and } P=V^{2} / R \text { or } P=I^{2} R \end{aligned} \] C1 \[ \begin{aligned} P=3.3 \times 10^{4} \times 3.0 \times 10^{7} \text { or } P=\left(3.0 \times 10^{7} \times 0.86\right) /\left(2.6 \times 10^{-5}\right) \text { or } P=\left(3.0 \times 10^{7}\right)^{2} / 910 \text { or } P=\left(3.3 \times 10^{4}\right)^{2} \times 910 P=9.9 \times 10^{11}(\mathrm{~W}) =990 \mathrm{GW} \end{aligned} \] A1
Question 6
6
An uncharged capacitor is connected in series with a battery, a switch and a resistor, as shown in Fig. 6.1. The battery has e.m.f. 9.0 V and negligible internal resistance. The capacitance of the capacitor is \(4700 \mu \mathrm{~F}\). The switch is closed at time t=0. During the time interval t=0 to \(t=4.0 \mathrm{~s}\), the charge passing through the resistor is 22 mC .
Question 6(a)
6(a)
Question 6(a)(i)
6(a)(i)
Calculate the energy transfer in the battery during the time interval t=0 to \(t=4.0 \mathrm{~s}\). energy transfer = J
Answer
energy =E Q C1
Question 4
4
A circuit used to measure the power transfer from a battery is shown in Fig. 4.1. The power is transferred to a variable resistor of resistance R. The battery has an electromotive force (e.m.f.) E and an internal resistance r. There is a potential difference (p.d.) V across R. The current in the circuit is I.
Question 4(d)
4(d)
Question 4(d)(i)
4(d)(i)
Using data from Fig. 4.2, calculate the power transferred to R for a current of 1.6 A . power = W
Answer
P=V I C1 \(P=2.9 \times 1.6=4.6(4.64) \mathrm{W} \quad\) A1
Question 4
4
Question 4(b)
4(b)
The supply voltage is changed to 110 V .
Question 4(b)(i)
4(b)(i)
Calculate the power output of the heater at this voltage, assuming there is no change in the resistance of the wire. power = W
Answer
P=625,620 or 630 W
Question 6
6
Two resistors A and B have resistances \(R_{1}\) and \(R_{2}\) respectively. The resistors are connected in series with a battery, as shown in Fig. 6.1. The battery has electromotive force (e.m.f.) E and zero internal resistance.
Question 6(b)
6(b)
The current in the circuit is I. State the rate of energy transformation in
Question 6(b)(ii)
6(b)(ii)
the resistor A .
Answer
\(\left(P_{\mathrm{R}}=\right) I^{2} R_{1}\) A1 [1]
Question 6(c)
6(c)
The resistors are made from metal wires. Data for the resistors are given in Fig. 6.2. Use information from Fig. 6.2 to determine the ratio power dissipated in A power dissipated in B . ratio =
Answer
\(R=\rho l / A\) or clear from the following equation B1 \(\begin{aligned} \text { ratio }=I^{2} R_{1} / I^{2} R_{2} & =\frac{\rho l / \pi d^{2}}{\rho(2 l) / \pi(2 d)^{2}} \text { or } R_{1} \text { has } 8 \times \text { resistance of } R_{2} & & \text { C1 } \\ & =8 \text { or } 8: 1 & & \text { A1 }\end{aligned}\) [3]
Question 6(d)
6(d)
The resistors A and B are connected in parallel across the same battery of e.m.f. E. Determine the ratio power dissipated in A power dissipated in B . ratio =
Answer
\(P=V^{2} / R\) or \(E^{2} / R \quad\) C1 ( V or E the same) hence ratio is 1 / 8 or 1: 8=0.125 (allow ecf from (c)) A1 [2]
Question 6
6
An electric heater is to be made from nichrome wire. Nichrome has a resistivity of \(1.0 \times 10^{-6} \Omega \mathrm{~m}\) at the operating temperature of the heater. The heater is to have a power dissipation of 60 W when the potential difference across its terminals is 12 V .
Question 6(c)
6(c)
A second heater, also designed to operate from a 12 V supply, is constructed using the same nichrome wire but using half the length of that calculated in (b). Explain quantitatively the effect of this change in length of wire on the power of the heater.
Answer
either current is doubled or power \(\propto 1 / R\) ………… ………………………… M1 power is doubled \(\ldots\).......................................................................... A1