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The values of current (I) flowing through a given resistor of resistance (R), for the corresponding values of potential difference (V) across the resistor are as given below :
V (Volts) | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 4.0 | 5.0 |
I (Amperes) | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.8 | 1.0 |
The plot between current $(I)$ and potential difference $(V)$ is given as-
Resistance of the resistor (R) = Slope of the above graph
Choose any two points P and Q on the graph,
Resistnce of the resistor (R) = Slope of line $PQ=\frac{4-1}{0.8-02}=\frac{3}{0.6}=\frac{30}{6}=5\Omega $
Alternate Solution
We know that-
$R=\frac{\Delta V}{\Delta I}$
Where,
$R$ = Resistance
$V$ = Potential difference or Voltage
$I$ = Current
$\Delta $ = shows change
Putting the given values we get-
$R=\frac{3-2}{0.6-0.4}$
$R=\frac{1}{0.2}$
$R=\frac{10}{2}$
$R=5\Omega $
NOTE: For every point on the graph, the resistance will remain the same.
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