Question 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

Question 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- Hosting Precise

Posted on 4th Sep 2021 02:55:45 in

Question 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.

Ans:

The voltage is to be plotted on x-axis and current on y-axis.

The graph will look like:-

Using Ohm’s law:-

$R = \frac{V}{I}$

where R = resistance, I = current and V = potential difference

So, $R = \frac{(V_{2} - V_{1})}{(I_{2} - I_{1})}$

Resistance is given by the slope of the graph.

Therefore using values,

$V_{2} = 3.4 V$

$V_{1} = 1.6 V$

$I_{2} = 1 A$

$I_{1} = 0.5 A$

$R = \frac{(3.4 - 1.6)}{(1 - 0.5)}$

R =3.6 Ω

$R = \frac{(V_{3} - V_{2})}{(I_{3} - I_{2})}$

$= \frac{(6.7 - 3.4)}{(2 - 1)} = 3.3 Ω$

$R = \frac{(V_{4} - V_{3})}{(I_{4} - I_{3})}$

$= \frac{(10.2 - 6.7)}{(3 - 2)} = 3.5 Ω$

$R = \frac{(V_{5} - V_{4})}{(I_{5} - I_{4})}$

$= \frac{(13.2 - 10.2)}{(4 - 3)} = 3 Ω$

As R is variable, therefore taking average of the resistances,

${R_{a v}}_{} = \frac{(3.6 + 3.3 + 3.5 + 3)}{4}$

$= \frac{13.4}{4} = 3.35 Ω$

This shows, resistance = 3.35Ω

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