(a) You are provided with a battery, an ammeter, a voltmeter, a resistance box, a key and connection wires.
(i) Set up circuit as shown in the diagram above.
(ii) With the key opened, measure and record the e.m.f. E\(_o\) of the battery
(iii) With the key closed, select the resistance R=1 on the resistance box. Read and record the current, 1Ω.
(iv) Evaluate I\(^{-1}\).
(v) Repeat the procedure for five other values of R=2Ω,3Ω, 4Ω, 5Ω, and 6Ω.
In each case, record I and evaluate I\(^{-1}\) results.
(vii) Plot a graph with R on the vertical axis and I\(^{-1}\) on the horizontal axis.
(Viii) Determine the slope, s, of the graph.
(ix) Determine the intercept C, On the vertical axis.
(x) State two precautions taken to ensure accurate results.
(b)(i) Define potential difference in an electric field.
(ii) A piece of resistance wire of diameter 0.2 mm and length 25 cm has a resistance of 7Ω. Calculate the resistivity of the wire of the battery. [ π = 22\7]
(vii) Slope = \(\frac{Δd}{ΔI^{-1}}\) = \(\frac{d_2 - d_1}{I_2 - I^{-1}\)
→ \(\frac{80 - 30}{3.333 - 15}\} = \(\frac{50}{1.833}\)
Slope = 27.2cm / A
Precaution:
Diameter d = 0.6cm = 0.006m
: r = \(\frac{0.006}{2}\) = 0.003m
Resistivity = 1.0 x 10\(^{-6}\)
R = 4Ω
R = \(\frac{ρL}{A}|) → L = \(\frac{RA}{ρ}\)
L = \(\frac{ 4 x 22 x 0.0000009}{7 x 1.0 x 10^{-6}}\)
L = 6.289 x 2 x 9 x 10\(^{-6}\) x 10\(^{-6}\)
L = 113.14 x 10\(^{-6 + 6}\)
L = 113.14
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}