NCERT Solutions
Class 10 Science
Electricity

Q.7
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –
I (amperes) 0.5 1.0 2.0 3.0 4.0
V (volts) 1.6 3.4 6.7 10.2 13.2
Plot a graph between V and I and calculate the resistance of that resistor.
The voltage is to be plotted on x-axis and current on y-axis.
The graph will look like:-
Using Ohm’s law:-
R = (V/I) where R = resistance, I = current and V = potential difference
So, R = (V2-V1)/ (I2-I1); Resistance is given by the slope of the graph.
Therefore using values, V2 = 3.4V, V1 = 1.6V, I2 = 1A and I1 =0.5A
So, R = (3.4-1.6)/ (1-0.5)
R =3.6 Ω
Similarly, R = (V3-V2)/ (I3-I2);
= (6.7 – 3.4) / (2 -1) = 3.3 Ω
Also R = (V4-V3)/ (I4-I3);
= (10.2 -6.7)/ (3-2) = 3.5 Ω
R= (V₅ - V₄)/ (I₅ - I₄) = (13.2 - 10.2)/ (4-3) = 3 Ω
As R is variable, therefore taking average of the resistances,
Rav = (3.6 + 3.3 + 3.5 + 3)/4 = (13.4/4) = 3.35Ω
This shows, resistance = 3.35Ω