Question 1 Report
Fig. 1.1 shows a circuit with a variable resistor Rv in series with a lamp L and a cell providing 6.0 V. An ammeter measures current and a voltmeter measures the p.d. across the lamp.
The student adjusts Rv and records: when Rv = maximum, I = 0.05 A, V(L) = 0.5 V; when Rv = 0, I = 0.50 A, V(L) = 5.8 V.
(a) Record the ammeter reading when Rv = maximum. [1]
(b) Calculate the total resistance in the circuit when Rv = maximum. [1]
(c) Calculate the resistance of the lamp at I = 0.05 A. [1]
(d) Calculate the resistance of the lamp at I = 0.50 A. [1]
(e) Explain why the lamp resistance is different at the two settings. [1]
(f) Calculate the power dissipated in the lamp at each setting. [2]
(g) State the range of voltage that the student can obtain across the lamp using this circuit. [1]
(h) Suggest one advantage of using a variable resistor rather than simply changing the supply voltage. [1]
(i) Plan an experiment to plot the full V-I characteristic of the lamp. State what the student should do and what measurements to record. [2]
(a) When \( R_v \) is at maximum, the ammeter reads:
\( I = 0.05 \text{ A} \) [1]
(b) Total resistance when \( R_v \) is at maximum:
\[ R_{\text{total}} = \frac{V_{\text{supply}}}{I} = \frac{6.0}{0.05} = 120 \; \Omega \][1]
This total includes both the variable resistor and the lamp in series.
(c) Resistance of the lamp at \( I = 0.05 \text{ A} \):
\[ R_{\text{lamp}} = \frac{V_L}{I} = \frac{0.5}{0.05} = 10 \; \Omega \][1]
(d) Resistance of the lamp at \( I = 0.50 \text{ A} \):
\[ R_{\text{lamp}} = \frac{V_L}{I} = \frac{5.8}{0.50} = 11.6 \; \Omega \][1]
(e) The lamp resistance is higher at the greater current (11.6 Ω vs 10 Ω) because the larger current heats the filament to a higher temperature. In metals, increased temperature causes lattice ions to vibrate more vigorously, increasing collisions with conduction electrons and raising the resistance. [1]
(f) Power dissipated in the lamp at each setting using \( P = IV \):
At \( R_v \) = maximum:
\[ P = 0.05 \times 0.5 = 0.025 \text{ W} \][1]
At \( R_v = 0 \):
\[ P = 0.50 \times 5.8 = 2.9 \text{ W} \][1]
The lamp is far brighter at the lower resistance setting because it dissipates over 100 times more power.
(g) The voltage across the lamp ranges from 0.5 V to 5.8 V. [1]
The variable resistor cannot produce exactly 0 V across the lamp (some current always flows), and the terminal p.d. cannot reach the full 6.0 V because of the cell's internal resistance.
(h) A variable resistor provides continuous, fine control over the current and voltage in the circuit without needing multiple power supplies or component swaps. A single adjustment sweeps through the full range smoothly. [1]
(i) Plan for plotting the full V-I characteristic: [2]
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