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 cellis given as

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

Answer 1

Answer: (D)

The electromotive force (e) or e.m.f. is the energy provided by a cell or battery per coulomb of charge passing through it, it is measured in volts (V). It is equal to the potential difference across the terminals of the cell when no current is flowing.
E = I(R+r)
E = 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.
That is, E = IR+Ir, E = V + Ir.
Rearranging the equation we get, $r=\frac{E-V}{R}$.
Substituting I=V/R in the above equation, we get, $r=\frac{E-V}{\frac{V}{R}}=\frac{(E-V)R}{V}$.
Hence, internal resistance $r=\frac{(E-V)R}{V}$.