Question 4
4
A battery of electromotive force 12 V and negligible internal resistance is connected to two resistors and a light-dependent resistor (LDR), as shown in Fig. 4.1. An ammeter is connected in series with the battery. The LDR and switch S are connected across the points XY .
structured5 marks
Question 4(b)
4(b)
The switch S is closed. The resistance of the LDR is \(4.0 \mathrm{k} \Omega\). Calculate the current in the ammeter. current = A
Mediumstructured3 marks
Answer
parallel resistance \(=3(\mathrm{k} \Omega) \quad\) C1 total resistance \(8+3=11(\mathrm{k} \Omega)\) C1 current \(=12 / 11 \times 10^{3}=1.09 \times 10^{-3}\) or \(1.1 \times 10^{-3} \mathrm{~A} \quad\) A1 [3]
Question 4(c)
4(c)
The switch S remains closed. The intensity of the light on the LDR is increased. State and explain the change to
structured2 marks
Question 4(c)(i)
4(c)(i)
the ammeter reading,
Mediumstructured2 marks
Answer
LDR resistance decreases M1 total resistance (of circuit) is less hence current increases A1
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.
structured1 marks
Question 4(b)
4(b)
Using Kirchhoff's second law, determine an expression for the current I in the circuit.
Mediumstructured1 marks
Answer
E=I(R+r) or \(I=E /(R+r) \quad\) (any subject) B1 [1]
Question 6
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.
structured5 marks
Question 6(c)
6(c)
For \(R=6.0 \Omega\),
structured5 marks
Question 6(c)(i)
6(c)(i)
show that the p.d. between points A and B is 9.6 V ,
Mediumstructured2 marks
Answer
current \(=2.4(\mathrm{~A})\) C1 p.d. across \(\mathrm{AB}=24-2.4 \times 6=9.6 \mathrm{~V}\) M1 or total resistance \(=10 \Omega \quad(=24 \mathrm{~V} / 2.4 \mathrm{~A}) \quad\) C1 (parallel resistance \(=4 \Omega\) ), p.d. \(=24 \times(4 / 10)=9.6 \mathrm{~V} \quad\) M1
Question 6(c)(ii)
6(c)(ii)
calculate the resistance of X ,
Mediumstructured3 marks
Answer
\(R(\mathrm{AB})=9.6 / 2.4=4.0 \Omega \quad \mathrm{C} 1\) 1 / 6+1 / X=1 / 4 [must correctly substitute for R ] C1 \(X=12 \Omega\) A1 or \(I_{\mathrm{R}}=9.6 / 6.0=1.6(\mathrm{~A})\) \(I_{\mathrm{X}}=2.4-1.6=0.8(\mathrm{~A})\) \(X(=9.6 / 0.8)=12 \Omega\)
Question 5
5
Fig. 5.1 shows a 12 V power supply with negligible internal resistance connected to a uniform metal wire AB . The wire has length 1.00 m and resistance \(10 \Omega\). Two resistors of resistance \(4.0 \Omega\) and \(2.0 \Omega\) are connected in series across the wire. Currents \(I_{1}, I_{2}\) and \(I_{3}\) in the circuit are as shown in Fig. 5.1.
structured1 marks
Question 5(a)
5(a)
1 marks
Question 5(a)(i)
5(a)(i)
Use Kirchhoff's first law to state a relationship between \(I_{1}, I_{2}\) and \(I_{3}\).
Easystructured1 marks
Answer
\(I_{1}=I_{2}+I_{3} \quad\) B1 [1]
Question 5(a)(ii)
5(a)(ii)
Calculate \(I_{1}\).
Mediumstructured0 marks
Answer
I=V / R or \(I_{2}=12 / 10(=1.2 \mathrm{~A}) \quad\) C1 \(R=[1 / 6+1 / 10]^{-1}[\) total \(R=3.75 \Omega]\) or \(I_{3}=12 / 6 \quad(=2.0 \mathrm{~A}) \quad\) C1 \(I_{1}=12 / 3.75=3.2 \mathrm{~A} \quad\) or \(I_{1}=1.2+2.0=3.2 \mathrm{~A} \quad\) A1 [3]
Question 5(a)(iii)
5(a)(iii)
Calculate the ratio x, where
Hardstructured0 marks
Answer
power =V I or \(I^{2} R\) or \(V^{2} / R\) \(x=\frac{\text { power in wire }}{\text { power in series resistors }}=\frac{I_{2}^{2} R_{\mathrm{w}}}{I_{3}^{2} R_{\mathrm{s}}}\) or \(\frac{V_{2}}{V_{3}}\) or \(\frac{V^{2} / R_{\mathrm{w}}}{V^{2} / R_{\mathrm{s}}} \quad\) C1 \(x=12 \times 1.2 / 12 \times 2.0=0.6(0)\) allow 3 / 5 or 3: 5 A1
Question 5
5
A uniform resistance wire A B has length 50 cm and diameter 0.36 mm . The resistivity of the metal of the wire is \(5.1 \times 10^{-7} \Omega \mathrm{~m}\).
structured0 marks
Question 5(b)
5(b)
The wire A B is connected in series with a power supply E and a resistor R as shown in Fig. 5.1. The electromotive force (e.m.f.) of E is 6.0 V and its internal resistance is negligible. The resistance of R is \(2.5 \Omega\). A second uniform wire C D is connected across the terminals of E. The wire C D has length 100 cm , diameter 0.18 mm and is made of the same metal as wire A B. Calculate
structured0 marks
Question 5(b)(i)
5(b)(i)
the current supplied by E,
Mediumstructured0 marks
Answer
resistance of \(\mathrm{CD}=8 \times\) resistance of \(\mathrm{AB}=20(\Omega) \quad \mathrm{C} 1\) current \(=V / R=6.0 / 4.0 \quad\) C1 [4]