Question bank
Practice A-Level CAIE Physics 10 1 Practical Circuits questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.
Question 5
A power supply of electromotive force (e.m.f.) 50 V and negligible internal resistance is connected in series with resistors of resistance \(100 \Omega\) and \(5 \Omega\), as shown. A voltmeter measures the potential difference (p.d.) across the \(5 \Omega\) resistor and an ammeter measures the current in the circuit. What are suitable ranges for the ammeter and for the voltmeter?
D
Question 2
Question 2(b)
Question 2(b)(i)
Draw a circuit diagram of the apparatus that could be used to make these measurements.
metal wire in series with power supply and ammeter B1 voltmeter in parallel with metal wire B1 rheostat in series with power supply or potential divider arrangement or variable power supply B1
Question 5
Question 5(a)
Explain why the terminal potential difference (p.d.) of a cell with internal resistance may be less than the electromotive force (e.m.f.) of the cell.
lost volts/energy used within the cell/internal resistance B1 when cell supplies a current B1
Question 5(b)
A battery of e.m.f. 4.5 V and internal resistance r is connected in series with a resistor of resistance \(6.0 \Omega\), as shown in Fig. 5.1. The current I in the circuit is 0.65 A . Determine
Question 5(b)(i)
the internal resistance r of the battery,
\(E=I(R+r) \quad\) C1 4.5=0.65(6.0+r) \(r=0.92 \Omega\) A1 [2]
Question 5(b)(ii)
the terminal p.d. of the battery, p.d. = V
I=0.65 (A) and \(V=I R \quad\) C1 \(V=0.65 \times 6=3.9 \mathrm{~V}\) A1 [2]
Question 5(c)
A second resistor of resistance \(20 \Omega\) is connected in parallel with the \(6.0 \Omega\) resistor in Fig. 5.1. Describe and explain qualitatively the change in the heating effect within the battery.
(circuit) resistance decreases B1 current increases M1 more heating effect A1 [3]
Question 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(a)
By reference to the circuit shown in Fig. 4.1, distinguish between the definitions of e.m.f. and p.d.
e.m.f. = chemical energy to electrical energy M1 p.d. = electrical energy to thermal energy M1 idea of per unit charge A1 [3]
Question 4(c)
The variation with current I of the p.d. V across R is shown in Fig. 4.2. Use Fig. 4.2 to determine
Question 4(c)(i)
the e.m.f. E,
\(E=5.8 \mathrm{~V}\) B1 [1]
Question 4(c)(ii)
the internal resistance r.
evidence of gradient calculation or calculation with values from graph e.g. \(5.8=4+1.0 \times r\) C1 \(r=1.8 \Omega\) A1 [2]
Question 4
Question 4(a)
Distinguish between potential difference (p.d.) and electromotive force (e.m.f.) in terms of energy transformations.
p.d. = energy transformed from electrical to other forms unit charge B1 e.m.f. = energy transformed from other forms to electrical unit charge B1 [2]
Question 4(b)
Two cells A and B are connected in series with a resistor R of resistance \(5.5 \Omega\), as shown in Fig. 4.1. Cell A has e.m.f. 4.4 V and internal resistance \(2.3 \Omega\). Cell B has e.m.f. 2.1 V and internal resistance \(1.8 \Omega\).
Question 4(b)(iii)
On Fig. 4.1, draw an arrow to show the direction of the current in the circuit. Label this arrow I.
arrow (labelled) I shown anticlockwise A1 [1]
Question 4(b)(iv)
Calculate 1. the p.d. across resistor R, 2. the terminal p.d. across cell A, 3. the terminal p.d. across cell B.
1. \(V=I \times R=0.24 \times 5.5=1.3(2) \mathrm{V}\) A1 [1] 2. \(V_{\mathrm{A}}=4.4-(I \times 2.3)=3.8(5) \mathrm{V}\) A1 3. either \(V_{\mathrm{B}}=2.1+(I \times 1.8)\) or \(V_{\mathrm{B}}=3.8-1.3\) C1 =2.5(3) V A1 [2]
Question 5
Question 5(b)
A potential divider circuit is shown in Fig. 5.2. The battery of electromotive force (e.m.f.) 12 V and negligible internal resistance is connected in series with resistors X and Y and thermistor Z. The resistance of Y is \(15 \mathrm{k} \Omega\) and the resistance of Z at a particular temperature is \(3.0 \mathrm{k} \Omega\). The potential difference (p.d.) across Y is 8.0 V .
Question 5(b)(i)
Explain why the power transformed in the battery equals the total power transformed in X, Y and Z.
(no energy lost in battery because) no/negligible internal resistance B1 [1]
Question 6
A battery is connected in series with resistors X and Y, as shown in Fig. 6.1. The resistance of X is constant. The resistance of Y is \(6.0 \Omega\). The battery has electromotive force (e.m.f.) 24 V and zero internal resistance. A variable resistor of resistance R is connected in parallel with X. The current I from the battery is changed by varying R from \(5.0 \Omega\) to \(20 \Omega\). The variation with R of I is shown in Fig. 6.2.
Question 6(a)
Explain why the potential difference (p.d.) between points A and C is 24 V for all values of R.
there are no lost volts/energy lost in the battery or there are no lost volts/energy lost in the internal resistance B1 [1]
Question 6(b)
Use Fig. 6.2 to state and explain the variation of the p.d. across resistor Y as R is increased. Numerical values are not required.
the current/ I decreases (as R increases) M1 p.d. decreases (as R increases) A1 or the parallel resistance (of X and R ) increases M1 p.d. across parallel resistors increases, so p.d. (across Y) decreases A1 [2]
Question 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(a)
State the energy transformation that occurs in
Question 6(a)(i)
the battery,
chemical to electrical B1 [1]
Question 6(a)(ii)
the resistors.
electrical to thermal / heat or heat and light B1 [1]
Question 6(b)
The current in the circuit is I. State the rate of energy transformation in
Question 6(b)(i)
the battery,
( \(\left.P_{\mathrm{B}}=\right) E I\) or \(I^{2}\left(R_{1}+R_{2}\right) \quad\) A1 [1]
Question 5
A student sets up a circuit with a battery, an ammeter, a heater and a light-dependent resistor (LDR) all in series. The battery has negligible internal resistance. A voltmeter is connected across (in parallel with) the heater.
Question 5(a)
On Fig. 5.1, complete the circuit diagram of this arrangement.
correct symbol for the heater or for the LDR M1 all correct symbols in series (ignore voltmeter) and no extra symbols A1 correct symbol for voltmeter and in parallel with the heater B1
Question 6
Question 6(c)
The electromotive force (e.m.f.) of the cell in Fig. 6.1 is 1.50 V . The values of \(R_{1}\) and \(R_{2}\) are \(10 \Omega\) and \(15 \Omega\) respectively. The terminal p.d. of the cell is 1.35 V . Calculate the internal resistance r of the cell.
current in circuit \(=1.35 /(10+15)(=0.054 \mathrm{~A})\) C1 r=(E-V) / I C1 \[ \begin{aligned} =(1.5-1.35) / 0.054 =2.8 \Omega \end{aligned} \] A1 or by potential divider principle \(\frac{0.15}{1.35}=\frac{r}{25}\) (C2) \(r=2.8 \Omega\) (A1) or \(I=1.35 /(10+15)(=0.054 \mathrm{~A})\) (C1) total resistance \(=1.50 / 0.054(=27.8 \Omega)\) r=27.8-25 (C1) \(r=2.8 \Omega\) (A1)
Question 6(d)
A resistor of resistance \(R_{3}\) is added to the circuit in Fig. 6.1, so that the circuit is as shown in Fig. 6.2. State and explain the effect, if any, of this change on:
Question 6(d)(i)
the current in the cell
the (total) resistance (of the circuit) has decreased (and e.m.f. is unchanged) M1 (the current (in the cell) will) increase A1
Question 6(d)(ii)
the terminal p.d. of the cell.
(as the current is greater and so there is a) larger p.d. across the internal resistance M1 (terminal p.d. will) decrease A1