Question

A battery of e.m.f E and internal resistance r is connected across a resistance R. The resistance R can be adjusted to any value greater than or equal to zero. A graph is plotted between the current (I) passing through the resistance and potential difference (V) across it. Select the correct alternatives.

• A. Internal resistance of the battery is $5\Omega$
• B. E.m.f. of the battery is 10 V.
• C. Maximum current which can be taken from the battery is 2 A.
• D. V – I graph can never be a straight line as shown in the figure.

Answer 1

Answer: (A), (B), (C)

Maximum current which can be taken from the battery is 2 A.

The circuit diagram for the given case can be drawn as shown below.
where $V$ is potential across battery, $E$ is emf of the battery, $r$ is internal resistance of the battery and $R$ is external resistance.
The relation between emf and potential across battery is given by
$\begin{array}{}V=E-Ir& \text{(1)}\end{array}$
The above relation matches with straight line equation, $y=mx+C$
where slope, $m=-r$ and y-intercept, $C=E$
Hence option D is incorrect.
From graph, y intercept is 10 V.
Since y-intercept, $C=E=10V$
Hence option B is correct.
Now slope of the $V-I$ graph is $m=\frac{0-10}{2-0}=-5$
Since slope, $m=-r$
$\Rightarrow r=5\Omega$
Hence option A is correct.
Maximum current can be drawn from the battery when external resistance $R$ is zero.
Thus relation (1) becomes,
$0=E-Ir$
$\Rightarrow E=Ir$
$I=\frac{E}{r}=\frac{10}{5}=2A$
Hence option C is correct.