Question

A cell of emf 'E' is connected across a resistance R. The potential difference between the terminals of the cell is found to be V. The internal resistance of the cell must be

• A. $\frac{2(E-V)V}{R}$
• B. $\frac{2(E-V)R}{E}$
• C. $\frac{(E-V)R}{V}$
• D. $(E-V)R$

Answer 1

The correct option is C $\frac{(E-V)R}{V}$

The electromotive force (e) or e.m.f. is the energy provided by a cell or battery per coulomb of charge passing through it, is measured in volts (V). It is equal to the potential difference across the terminals of the cell when no current is flowing.
$\epsilon =\frac{E}{Q}$
where,
$\epsilon$ = electromotive force in volts, V
E = energy in joules, J
Q = charge in coulombs, C
Batteries and cells have an internal resistance (r) which is measured in ohms. When electricity flows round a circuit the internal resistance of the cell itself resists the flow of current and so thermal (heat) energy is wasted in the cell itself.
$\epsilon =I(R+r)$
where,
$\epsilon$ = electromotive force in volts, V
I = current in amperes, A
R = resistance of the load in the circuit in ohms,
r = internal resistance of the cell in ohms
The above equation can be written as $\epsilon =IR+Irthatis,\epsilon =V+Ir$.
The resistance of the cell is given as $Ir=\epsilon -Vthatis,r=\frac{\epsilon -V}{I}$ - eqn 1
and the resistance of the circuit is given as $R=\frac{V}{I}$ - eqn 2.
Dividing eqn 1 and eqn 2, we get
$\frac{r}{R}=\frac{\epsilon -V}{V}.Thatis,r=\frac{R(\epsilon -V)}{V}$
Hence, The internal resistance of the cell must be $r=\frac{R(\epsilon -V)}{V}$.