A-LevelPhysicsD.C. circuits9702/21 · Oct/Nov 2021 · Q511 Marks
(b) The circuit shown in [Fig. 5.1] contains a battery of electromotive force (e.m.f.) E and negligible internal resistance connected to four resistors R1, R2, R3 and R4, each of resistance R. The current in R3 is 0.30A and the potential difference (p.d.) across R4 is 2.4V.
- (a)State Kirchhoff's first law.[2]
- (b)(i)Show that R is equal to 4.0Ω.[2]
- (b)(ii)Determine the e.m.f. E of the battery.[2]
- (c)The battery in (b) is replaced with another battery of the same e.m.f. E but with an internal resistance that is not negligible. State and explain the change, if any, in the total power produced by the battery.[2]
- (d)The resistors in the circuit of [Fig. 5.1] are made from nichrome wire of uniform radius 240 µm. The length of this wire needed to make each resistor is 0.67 m. Calculate the resistivity of nichrome.[3]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.
- [M1] sum of current(s) in = sum of current(s) out
- [A1] at a junction (in a circuit)
- [B1] (current in R₄ or R₁ =) 0.30 + 0.30
- [A1] (R =) 2.4/0.60 = 4.0 (Ω)
- [B1] (p.d. across R₃ or R₂ =) 2.4/2
- [A1] (R =) 1.2/0.30 = 4.0 (Ω)
- [C1] E = 2.4 + 2.4 + 1.2
- [A1] = 6.0 V
- [C1] total resistance = 10 (Ω)
- [A1] E = 10 x 0.60
- [B1] total resistance increases
- [B1] current decreases (in battery) so total power decreases
- [C1] resistivity = R A / L
- [C1] = 4.0 × π × (240 × 10⁻⁶)² / 0.67
- [A1] = 1.1 × 10⁻⁶ Ω m
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