This bicycle is fitted with two lamps so a rider can be seen at night. Fig. 1 shows the circuit planned by a technician. Each lamp is connected directly acr...
Question 1 Report
This bicycle is fitted with two lamps so a rider can be seen at night. Fig. 1 shows the circuit planned by a technician. Each lamp is connected directly across the 6.0 V battery. The technician chooses this arrangement because a rear lamp may fail during rain, but the front lamp should still receive energy from the battery. A fuse is also placed close to the positive terminal in the cable that carries current to both lamps.
(a) Which type of circuit connection is shown for the lamps? [1]
(b) What happens to the front lamp if the rear lamp breaks? [1]
(c) Calculate the current from the battery when each lamp has a resistance of 12 ohms. [2]
(d) Give one reason for fitting the fuse near the battery. [1]
This circuit is used by a repair shop to compare two metal wires for a model railway. Fig. 1 shows wire X and wire Y connected one at a time to the same 6.0 V supply. Both wires have equal length and diameter. The ammeter reading for X is 0.50 A, while the reading for Y is 0.20 A. The shop needs the wire that gives a larger resistance for a speed-control circuit.
(a) Calculate the resistance of wire X. [2]
(b) Calculate the resistance of wire Y. [1]
(c) Which wire should the shop choose for the speed-control circuit? [1]
(d) Give one variable that must be controlled for this comparison to be fair. [1]
Answer Details
Bicycle lamps
(a) The lamps are connected in parallel. [1]
(b) If the rear lamp breaks, the front lamp stays on because it remains on its own complete branch and continues to have 6.0 V across it. [1]
(c) For each lamp:
\[I=\frac{V}{R}=\frac{6.0\ \mathrm{V}}{12\ \Omega}=0.50\ \mathrm{A}\]
There are two parallel branches, so:
\[I_\text{total}=0.50\ \mathrm{A}+0.50\ \mathrm{A}=1.0\ \mathrm{A}\]
The battery current is 1.0 A. [2]
(d) The fuse protects the cable and circuit if an excessive current flows, for example during a fault. [1]
Comparing metal wires
(a) \[R_X=\frac{V}{I}=\frac{6.0\ \mathrm{V}}{0.50\ \mathrm{A}}=12\ \Omega\]
Wire X has resistance 12 \(\Omega\). [2]
(b) \[R_Y=\frac{6.0\ \mathrm{V}}{0.20\ \mathrm{A}}=30\ \Omega\]
Wire Y has resistance 30 \(\Omega\). [1]
(c) Choose wire Y, because 30 \(\Omega\) is greater than 12 \(\Omega\). [1]
(d) Control the temperature of the wires. Length and diameter are also acceptable, because each can affect resistance. [1]
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