You have been provided with an accumulator E, a standard resistor Rx, two resistance boxes R\(B_1\) and R\(B_2\), two keys \(K_1\) and \(K_2\) and other necessary apparatus.
Measure and record the e.m.f of the accumulator.
Connect a circuit as shown above.
Set the resistance R, in the resistance box such that R in R\(B_1\) = R in R\(B_2\) = 1\(\Omega\).
With \(K_1\) open and \(K_2\) closed, measure and record the potential difference V across the standard resistor Rx. (v) Close K, and K,. Read and record the potential difference \(V_o\) across Rx.
Evaluate V\(_{1}^{-1}\)
Repeat procedure (v) for four other values of R = 2, 3, 4 and 5\(\Omega\) respectively. In each case, ensure that the value of R in R\(B_1\) is equal to the value of R in R\(B_2\).
Evaluate V\(_{1}^{-1}\) in each case. Tabulate your readings.
Plot a graph of V\(_{1}^{-1}\) on the vertical axis against R on the horizontal axis starting both axes from the origin (0,0).
Determine the slope, s, of the graph and the intercept on the vertical axis.
Evaluate y = \(\frac{1}{s}\)
State two precautions taken to ensure accurate results
(b)i. Explain what is meant by the potential difference between two points in an electric circuit
ii. A cell has an e.m.f. of 3 V. When it is connected across a resistor of resistance 4\(\Omega\), a current 0.5A passes through the circuit. Calculate the internal resistance of the cell.
Verified Correct Answer: .
Physical Principles & Numerical Solution
Table of values (Observation)
R (\(\Omega\)
\(V_1\)(v)
V\(_{1}^{-1}\)
1
1.50
0.667
2
1.30
0.769
3
1.15
0.870
4
1.10
0.909
5
0.90
1.111
Slope (s) = \(\frac{\bigtriangleup {V}_{1}^{-1}}{DR} = \frac{1.22 - 0.6}{6.5 - 0.3}\)
= \(\frac{0.62}{6.20}\) = 0.1
Intercept (c) = 0.57
Evaluation: y = \(\frac{1}{S} = \frac{1}{0.1}\) = 10
Precautions:
I removed the key when not taking readings
I ensured tight connections/clean terminals
I avoided parallax error in reading the voltmeter
I corrected for zero error on the voltmeter
(b)i. The potential difference (pd) (in volts) between two points in a circuit is the work done (in joules) when one unit of positive charge (coulomb) moves from one point to the other
ii. E = IR + Ir, 3 =0.5 x 4 + 0.5r, 3 = 2 +0.5r
r = \(\frac{3 - 2}{0.5}\) = 2\(\Omega\)
Source: Myschool Past Questions Archive (Record #60043)