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:
Answer & explanation
Correct answer: option 2
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
$4-R^2=0$
$(2-R)(2+R)=0$
$R=2Ω$
$∴I=\frac{12}{2+2}=3A$