A battery has an e.m.f 12 V and internal resistance $2 \Omega$. This battery is connected to a variable resistor 0 to $5 \Omega$. The value of current for which the rate of heat production in the resistor is maximum:

3 A

The correct answer is Option (2) → 3 A

Rate of heat production (power dissipation) in resistor, P is :

$P = I^2 R$

where,

I = Current in the circuit

R, Variable Resistor = 0 Ω - 5 Ω

r, Internal resistance of battery = 2 Ω

$R_{total}=r+R$ [As they are in series]

By ohm's law,

$I=\frac{ε}{R_{total}}=\frac{12}{r+R}$

$∴P_R=\left(\frac{12}{2+R}\right)^2R$

$=\frac{144R}{(2+R)^2}$

$\frac{dP_R}{dt}=\frac{144(2+R)^2-144R.2(2+R)}{(2+R)^4}$

$=\frac{144[4-R^2]}{(2+R)^4}$

Set the derivative equal to zero to find critical point

$4-R^2=0$

$(2-R)(2+R)=0$

$R=2Ω$

$∴I=\frac{12}{2+2}=3A$