Loading....

WAEC SSCE - Physics - 2007 (Practical)

Question 1 Report

You have been provided with an accumulator E, a standard resistor Rx, two resistance boxes RB\(_{1}\) and RB\(_{2}\), two keys K\(_{1}\) and K\(_{2}\) and other necessary apparatus.

  1. Measure and record the e.m.f of the accumulator.
  2. Connect a circuit as shown above.
  3. Set the resistance R, in the resistance box such that R in RB\(_{1}\) = R in RB\(_{2}\) = \(1\Omega\).
  4. 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.
  5. Evaluate V\(_{1}^{-1}\)
  6. 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 RB\(_{1}\) is equal to the value of R in RB\(_{2}\).
  7. Evaluate V\(_{1}^{-1}\) in each case. Tabulate your readings.
  8. 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).
  9. Determine the slope, s, of the graph and the intercept on the vertical axis.
  10. Evaluate \(y = \frac{1}{s}\)
  11. 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.

Answer Details

Measurement of e.m.f. of the accumulator

E.m.f. of the accumulator, \(E = 2.0\ \text{V}\).

Observation and table of readings

For each setting the resistance in \(RB_1\) was made equal to the resistance in \(RB_2\) (\(R\) in \(RB_1 = R\) in \(RB_2 = R\)). With \(K_1\) and \(K_2\) closed the potential difference \(V_1\) across the standard resistor \(R_x\) was read, and \(V_1^{-1}\) evaluated.

\(R\ (\Omega)\)\(V_1\ (\text{V})\)\(V_1^{-1}\ (\text{V}^{-1})\)
11.500.667
21.300.769
31.150.870
41.100.909
50.901.111

Graph of \(V_1^{-1}\) against \(R\)

The graph of \(V_1^{-1}\) (vertical axis) against \(R\) (horizontal axis), with both axes starting from the origin, is a rising straight line.

graph
Straight-line graph of V₁⁻¹ against R; slope s = 0.10 V⁻¹Ω⁻¹, vertical-axis intercept c = 0.57 V⁻¹.

Slope and intercept

Taking two widely spaced points on the line of best fit, \((0.3,\ 0.60)\) and \((6.5,\ 1.22)\):

\[ s = \frac{\Delta (V_1^{-1})}{\Delta R} = \frac{1.22 - 0.60}{6.5 - 0.3} = \frac{0.62}{6.2} = 0.10\ \text{V}^{-1}\,\Omega^{-1} \]

The intercept on the vertical axis (value of \(V_1^{-1}\) where the line cuts \(R = 0\)) is

\[ c = 0.57\ \text{V}^{-1}. \]

Evaluation of \(y = \dfrac{1}{s}\)

\[ y = \frac{1}{s} = \frac{1}{0.10} = 10\ \text{V}\,\Omega. \]

Two precautions

  1. I opened the keys when readings were not being taken, to avoid heating of the resistors and drift in the cell.
  2. I made clean, tight connections at all terminals and read the voltmeter with my eye directly in front of the pointer to avoid parallax error (also correcting for any zero error).

(b)(i) Potential difference between two points

The potential difference between two points in an electric circuit is the work done (in joules) in moving one coulomb of positive charge from one point to the other. It is measured in volts.

(b)(ii) Internal resistance of the cell

Given \(E = 3\ \text{V}\), \(R = 4\ \Omega\) and current \(I = 0.5\ \text{A}\):

\[ E = IR + Ir \] \[ 3 = (0.5 \times 4) + 0.5r \] \[ 3 = 2 + 0.5r \] \[ 0.5r = 1 \] \[ r = \frac{1}{0.5} = 2\ \Omega. \]
1
2
3