A-LevelPhysicsD.C. circuits9702/21 · Oct/Nov 2020 · Q78 Marks
The current in the circuit is I.
- (a)Define the ohm.[1]
- (b)A uniform wire has resistance 3.2 Ω. The wire has length 2.5 m and is made from metal of resistivity 460 nΩ m. Calculate the cross-sectional area of the wire.[3]
- (c)(i)A cell of electromotive force (e.m.f.) E and internal resistance r is connected to a variable resistor of resistance R, as shown in Fig. 7.1. [Figure 7.1] State, in terms of energy, why the potential difference across the variable resistor is less than the e.m.f. of the cell.[1]
- (c)(ii)State an expression for E in terms of I, R and r.[1]
- (c)(iii)The resistance R of the variable resistor is changed so that it is equal to r. Determine an expression, in terms of only E and r, for the power P dissipated in the variable resistor.[2]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.
- [B1] volt / ampere
- [C1] R = ρL / A
- [C1] A = 460 x 10-9 × 2.5 / 3.2
- [A1] = 3.6 x 10-7 m²
- [B1] energy is dissipated in the internal resistance/r
- [B1] E = IR + Ir or E = I (R + r)
- [C1] P = I²R or P = I²r
- [A1] I = E/2r (so) P = E² / 4r
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