Question

A cell of e.m.f. E and internal resistance r drives a current I through an external resistance R:

• A. the cell supplies EI power
• B. heat is produced in R at the rate EI
• C. heat is produced in R at the rate $EI\left(\frac{R}{R+r}\right)$
• D. heat is produced in the cell at the rate $EI\left(\frac{r}{R+r}\right)$

Answer 1

Answer: (A), (C), (D)

The correct options are
A the cell supplies EI power
C heat is produced in R at the rate $EI\left(\frac{R}{R+r}\right)$
D heat is produced in the cell at the rate $EI\left(\frac{r}{R+r}\right)$

If we consider a perfect battery with no internal resistance, then its emf would be equal to the potential difference across the terminals. But, batteries have internal resistance, the terminal voltage is not equal to the emf for a battery. If we consider a real battery like a perfect battery with a resistor in series (this resistor is the internal resistance). When we pass the supply, the potential increases by an amount E (the potential supplied by the perfect cell) and decreases by Ir (the voltage drop across the internal resistance. So, the potential difference across the real cell would be given by: V = E - Ir.
Now lets say there is a load resistor R connected to the circuit. The load resistor might be a simple resistive circuit element, or it could be the resistance of an electrical device connected to the battery. The potential difference across the load resistor is V = IR. Using $V=E-Ir$ we get: $IR=E-Ir$
and solving for emf we get:
$E=IR+Ir$
$E=I(R+r)$
Solving for current we get: I = E/(R + r).
The power or energy is given by $P=\left(I^2\right)\left(R\right)=\frac{\left(E^2\right)\left(R\right)}{(r+R{)}^2}$.
Hence, the options A, C and D are correct.