Question 1 Report
Fig. 1 is a circuit used by a museum technician before placing a low-power warming strip inside a display case. The strip keeps the air temperature above 10 degrees C so that water does not condense on a metal exhibit. A variable power supply provides 12 V. Students close the switch for 10 minutes and observe that the ammeter reading remains 0.50 A. The voltmeter is connected across the warming strip. The technician will fit a fuse in the supply lead before the circuit is used overnight.
(a) Calculate the resistance of the warming strip. Use the equation resistance = voltage / current. [2]
(b) Calculate the power transferred electrically to the strip. [2]
(c) Calculate the energy transferred by the strip in 10 minutes. Give your answer in J. [2]
(d) Describe the energy transfers that cause the air in the display case to become warmer. [2]
(e) Which fuse rating is most suitable: 0.25 A, 1 A or 13 A? [2]
(a) Use \(R=V/I\):
\[R=\frac{12\ \mathrm{V}}{0.50\ \mathrm{A}}=24\ \Omega\]
The warming strip has a resistance of 24 \(\Omega\). [2]
(b) Use \(P=VI\):
\[P=12\ \mathrm{V}\times0.50\ \mathrm{A}=6.0\ \mathrm{W}\]
The power transferred is 6.0 W. [2]
(c) \(10\) minutes \(=600\ \mathrm{s}\).
\[E=Pt=6.0\ \mathrm{W}\times600\ \mathrm{s}=3600\ \mathrm{J}\]
The energy transferred is 3600 J. [2]
(d) Electrical energy is transferred to thermal energy in the warming strip. Thermal energy is then transferred from the strip to the surrounding air and case, increasing their temperature. [2]
(e) A 1 A fuse is most suitable. It is above the normal current of 0.50 A, so it should not melt during normal use, but is far lower than 13 A and will melt if fault current becomes too large. [2]
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