You are provided with a battery of e.m.f. E, a key K, a voltmeter, a standard resistor \(R_0\) = 2\(\Omega\), a resistance box R, and some connecting wires.
i. Measure and record the e.m.f. E of the battery.
i. Set up a circuit as shown in the diagram above with the key open.
iii. Set the resistance on the resistance box to R 22.
iv. Close the key, read and record the potential difference V on the voltmeter.
v. Evaluate V\(^{-1}\)
vi. Repeat the procedures for five other values of R = 5\(\Omega\) 10\(\Omega\),12\(\Omega\),15\(\Omega\) and 20\(\Omega\). In each case, record V and evaluate V\(^{-1}\)
vii. Tabulate the results.
viii. Plot a graph with R on the vertical axis and V\(^{-1}\) l on the horizontal axis, starting both axes from the origin (0,0).
ix. Determine the slope, s, of the graph and the intercept c on the vertical axis.
x. Calculate \(\propto\) and \(\beta\) from the equations s = \(R_0\) \(\propto\) and c= - (\(R_0\)+B).
xi. State two precautions taken to obtain accurate results.
A
Connecting wires
B
Earth wire
C
Fuse
D
Switch
Verified Correct Answer: A.
Physical Principles & Numerical Solution
\(R_0\) = 2\(\Omega\), emf of the battery = 2.0V
Table of value
S/N
R\(\Omega\)
V
V
1
2
3
4
5
6
2
5
10
12
15
20
4.70
2.40
1.75
1.40
1.10
0.76
0.214
0.417
0.571
0.714
0.909
1.320
Slope = \(\frac{\bigtriangleup R}{\bigtriangleup V^{-1}}\) = \(\frac{30.00-0.00}{1.50-0.15}\)
= \(\frac{30.00}{1.35}\) = 22
Intercept on the horizontal axis = 0.15
Calculation of \(\propto\)
\(\propto\) = \(\frac{S}{R_0} = \frac{22}{2}\) = 11
Calculation of \(\beta\)
\(\beta\) = -(\(R_0\) + C) = - (2 + 0.15) = -(2.15) = -2.15
Precuations
- Key opened in between reading/key opened when readings were not taken.
- Tight connections ensured
- Avoided parallax error when taking reading on voltmetre
- Repeated readings shown on table
- Noted/corrected zero error on voltmetre
- Clean terminals ensured.
(b) i. R\(_{AB}\) = \(\frac{4 \times 5}{4+5} = \frac{20}{9}\)
P\(_{AB}\) =\(\frac{V_{AB^2}}{R_{AB}}\)
= \(\frac{4^2 \times9}{20}\) = 7.2 W
ii. I = \(\frac{p}{v}\)=\(\frac{3.6\times10^3}{240}\)
= 15A
The breaker will not open since the current drawn (15A) is less than (20A).
Source: Official Exam / Past Question Archive