A network of three resistors of resistances R1, R2 and R3 is shown in Fig. 6.1. [Figure 6.1] The individual currents in the resistors are I1, I2 and I3. The total current in the combination of resistors is I and the potential difference across the combination is V. A battery of electromotive force (e.m.f.) 8.0 V and internal resistance r is connected to three resistors X, Y and Z, as shown in Fig. 6.2. [Figure 6.2] Resistor Y has a resistance of 16 Ω. The current in resistor X is 0.49A and the current in resistor Y is 0.45A. Resistors X and Y in Fig. 6.2 are made from wires of the same material and cross-sectional area. The average drift speed of the free electrons in X is 2.1 × 10⁻⁴ m s⁻¹.
- (a)Show that the combined resistance R of the network is given by 1/R = 1/R₁ + 1/R₂ + 1/R₃[2]
- (b)(i)Calculate: the current in the battery current = .............................................. A[1]
- (b)(ii)the internal resistance r of the battery. r = .............................................. Ω[2]
- (c)Calculate the average drift speed v of the free electrons in Y. v = .............................................. ms⁻¹[2]
- (d)Resistor Z in Fig. 6.2 is replaced by a new resistor of smaller resistance. State and explain the effect, if any, on the terminal potential difference of the battery.[2]
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- [B1] I = I₁ + I₂ + I₃
- [B1] (V/ R) = (V/ R₁) + (V/ R₂) + (V/R₃) or (I / V) = (I₁ / V) + (I₂ / V) + (I₃ / V) and 1/ R = 1/R₁ + 1 / R₂ + 1 / R₃
- [A1] current = 0.49 + 0.45 = 0.94 A
- [C1] 8.0 = (0.94 × r) + (0.45 × 16)
- [A1] r = 0.85 Ω
- [C1] I = Anvq
- [A1] v = (0.45 / 0.49) × 2.1 × 10⁻⁴ = 1.9 × 10⁻⁴ m s⁻¹
- [B1] total/combined resistance decreases
- [B1] (current in battery increases so terminal) potential difference decreases
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