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.
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).
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