(a)(i) Describe, with the aid of a circuit diagram, an experiment to measure the resistance of a wire given an ammeter of low resistance, a battery, a key, ...
Question 1 Report
(a)(i) Describe, with the aid of a circuit diagram, an experiment to measure the resistance of a wire given an ammeter of low resistance, a battery, a key, a rheostat, va high-resistance voltmeter and some connecting wires.
(ii) State two precautions necessary to obtain an accurate result.
(b) Using the experimental result and any necessary measurements, explain how the resistivity of the wire may be determined.
(c) Two cells, each of e.m.f. 2V and internal resistance 0.552, are connected in series. They are made to supply current to a combination of three resistors, one of resistance 20 connected in series to a parallel combination of two other resistors each of resistance 3Q. Draw the circuit diagram and calculate the:
(i) current in the circuit
(ii) potential difference across the parallel combination of the resistors
(iii) lost volts of the battery.
Answer Details
(a)(i) Measurement of the resistance of the wire
Connect the battery, key K, rheostat, ammeter A and the test wire X in series. Connect the high-resistance voltmeter V in parallel across the test wire, as shown below.
Close the key and set the rheostat initially to give a small current. Record the current I from the ammeter and the potential difference V across the wire. Adjust the rheostat to obtain several other pairs of readings. Plot a graph of V against I. The gradient of the straight-line graph gives the resistance, since
\[V=IR\qquad\therefore\qquad R=\frac{\Delta V}{\Delta I}.\]
(a)(ii) Precautions
- Open the key when readings are not being taken, so that the wire does not heat up and change its resistance.
- Ensure that all electrical connections are clean and tight.
- Read the ammeter and voltmeter at eye level to avoid parallax error.
- Use small currents to minimise heating of the wire.
(b) Determination of resistivity
Measure the length \(L\) of the test wire with a metre rule. Measure its diameter \(d\) at several positions and in perpendicular directions using a micrometer screw gauge, and obtain the mean diameter. Its cross-sectional area is
\[A=\frac{\pi d^2}{4}.\]
If \(R\) is the resistance obtained from the gradient of the \(V\)-against-\(I\) graph, then
\[R=\frac{\rho L}{A}.\]
Hence the resistivity of the material of the wire is
\[\boxed{\rho=\frac{RA}{L}=\frac{R\pi d^2}{4L}}\]
in \(\Omega\,\text{m}\).
(c) Circuit and calculation
The two cells are in series, so
\[E=2+2=4.0\text{ V},\qquad r=0.5+0.5=1.0\ \Omega.\]
The equivalent resistance of the two \(3\ \Omega\) resistors in parallel is
\[R_p=\frac{3\times3}{3+3}=1.5\ \Omega.\]
Thus, the external resistance is
\[R_{\rm ext}=2+1.5=3.5\ \Omega,\]
and the total resistance, including internal resistance, is
\[R_T=3.5+1.0=4.5\ \Omega.\]
(i) Current in the circuit
\[I=\frac{E}{R_T}=\frac{4}{4.5}=\frac{8}{9}=\boxed{0.89\text{ A}}\]
(ii) Potential difference across the parallel combination
\[V_p=IR_p=\frac{8}{9}\times1.5=\frac{4}{3}=\boxed{1.33\text{ V}}\]
(iii) Lost volts of the battery
\[V_{\rm lost}=Ir=\frac{8}{9}\times1.0=\boxed{0.89\text{ V}}\]
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