Examstyle Questions: Kirchoff's Laws and Potential Dividers

For the circuit below, which answer gives the correct equations for Kirchhoff's second law?
 Loop 1: 3V=12I_{1}+6I_{2}, Loop 2: 5V=6 I_{1}8 I_{2}
 Loop 1: 5V=6 I_{1}+8 I_{2}, Loop 2: 3V=12I_{1}+6I_{2}
 Loop 1: 3V=12I_{1}6I_{2}, Loop 2: 5V=6 I_{1}+8 I_{2}
 Loop 1: 3V=12I_{1}+6I_{2}, Loop 2: 5V=6 I_{1}+8 I_{2}
 a) Kirchhoff's first law is based on the conservation of an electrical quantity. State the law and the quantity conserved.
(2 Marks)

a) State Kirchoff's first law.
(2 Marks)
b) Fig. 3.1 shows part of an electrical circuit.
i) Name the component marked X.
(1 Mark)
ii) Determine the magnitude of the currents I_{1}, I_{2}, and I_{3} .
I_{1} = ................ mA
I_{2} = ............... mA
I_{3} = ............... mA
(3 Marks)

Fig. 4.1 shows a simple design for a 'movement' sensor used in an earthquake region. The supply has negligible internal resistance.
A resistance wire is stretched between two rigid steel plates, not shown in the diagram. During an earthquake, ground movement changes the seperation between the plates and so the length of wire changes.
The wire has a radius of 0.62 mm and length 32 cm. It is made of a material of resistivity 6.8 x 10^{6} Ωm.
a) Show that the resistance of the wire is 1.8 Ω.
(3 Marks)
b) Calculatethe potential difference (p.d.) between A and B.
p.d. = ................ V
(3 Marks)
c) The length of the wire increases. State and explain the effect on the p.d. between A and B.
(2 Marks)

a) Sum of the currents = zero (at junction)OR
sum of the currents in = sum of currents out (at junction)
(2 Marks)

a) sum of current(s) into a point / junction = sum of current(s) out (from the point / junction) (1 for ommission of 'point' or 'sum' in the statement of the law)
(Algebraic sum of current(s) at a point = 0)
(2 Marks)
b) i) Thermistor
(1 Mark)
ii) I_{1} = 51 mA
I_{2} = 9 mA
I_{3} = 29 mA
(3 Marks)

a)
(3 Marks)
b)
(3 Marks)
c) The p.d. increases because the resistance (of wire) increases (as it gets thinner / longer)
(2 Marks)
Fig 4.1 shows a potential divider circuit. The battery has negligible internal resistance and the voltmeter has very high resistance.
i) Show that the voltmeter reading is 1.5V.
(2 Marks)
b) An electric device rated at 1.5V, 0.1A is connected between the terminals X and Y. The device has constant resistance. The voltmeter reading drops to a very low value and the device fails to operate, even though the device itself is not faulty.
i) Calculate the total resistance of the device and the 400 Ω resistor in parallel.
Resistance =.......................................Ω
(3 Marks)
ii) Calculate the p.d. across the device when it is connected between X and Y.
p.d. =.........................................V
(2 Marks)
iii) Why does the device fail to operate?
(2 Marks)
(Marks available: 25)
Answer outline and marking scheme for question:
i) V = V_{0} x R_{2} / (R_{1} + R_{2})
V = 6.0 x 400 / (1200 + 400)
V = 1.5 (V)
{Answer of 4.5(V) scores 1/2)
OR
R = V/I and R = R_{1} + R_{2}
I = 6.0 / 1600 = 3.75 x 10^{3}(A)
V = 3.75 x 10^{3} x 400
V = 1.5(V)
(2 Marks)
b) i) RD = 1.5 / 0.1 = 15Ω
R = R_{1}R_{2} / (R_{1}+R_{2}) / 1R_{T}= Σ1/R_{1}
R = 400 x 15 / (400 + 15) = 14.5 = 15Ω (possible e.c.f)
(3 Marks)
ii) V ≈ 6.0 x 15 / (1200 + 15)
V ≈ 0.07V
{answer of 5.93 scores 1/2}
OR
I≈ 6.0 / 1200 + 15 ≈ 4.98 x 10^{3} (A)
V ≈ 4.98 x 10^{3} x 15 = 0.07 (V)
(2 Marks)
iii) Resistance of device is small(er) / current in device is small(er) / p.d. across device is small(er)
(2 Marks)
(Marks available: 25)