Question

A cell of 2V and internal resistance 1 ohm is connected to a conductor whose volt-ampere relation is given by ${V^2} = \alpha I$ where $\alpha$ is constant and is equal to 100.
Find the current drawn by the cell.
A. 10 mA
B. 40 mA
C. 50 mA
D. 20 mA

Answer 1

Hint: When a cell with an emf $E$ and internal resistance $r$ is connected to an external resistance $R$, the actual voltage realized on the external resistance is lesser than the emf. This voltage is called terminal voltage and the current that flows through the circuit is called terminal current.
The terminal (branch) current is given by the formula:
$I = \dfrac{E}{{R + r}}$

Complete step by step solution:
The Volt-ampere relation is given in the problem: ${V^2} = \alpha I$
If we rearrange the equation in the following way, we can obtain the expression in terms of external resistance $R$.
$
  {V^2} = \alpha I \\
  \dfrac{{{V^2}}}{I} = \alpha = 100 \\
  \dfrac{{V \times V}}{I} = 100 \\
  RV = 100 \\
  \because \dfrac{V}{I} = R \to Ohm'sLaw \\
  R = \dfrac{{100}}{V} \\
 $
Hence, for the value of voltage, $V= 2V$, the corresponding resistance $R$.
$R = \dfrac{{100}}{2} = 50\Omega $
The terminal (branch) current is given by the formula:
$I = \dfrac{E}{{R + r}}$
$
  E = 2V \\
  r = 1\Omega \\
  R = 50\Omega \\
 $
Substituting the values in the above equation, we get:
$ I = \dfrac{E}{{R + r}} $
$ \Rightarrow I = \dfrac{2}{{50 + 1}} = \dfrac{2}{{51}} = 0.039A \approx 0.04A $
$ \Rightarrow I = 40mA $

$\therefore$ The current drawn by the cell is $40mA$. Hence, the correct option is Option B.

Note:
The voltage available in the circuit will always be lesser than the emf because of the internal resistance. Hence, the terminal voltage $V$ is always lesser than the emf $E$. If in any scenario, we obtain the terminal voltage greater than the emf $E$, it is evident that you are on the wrong path to the solution and you should check the steps again.