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
The heat produced per second in the resistor is \( P = I^{2}R \), where the current is fixed by the whole series circuit, \( I = \frac{\varepsilon}{R + r} \). The power in R is largest when the external resistance equals the internal resistance (maximum power transfer).
\( \displaystyle P = \frac{\varepsilon^{2}R}{(R + r)^{2}} \)
Since \( (R + r)^{2} = (R - r)^{2} + 4Rr \), this can be written as
\( \displaystyle P = \frac{\varepsilon^{2}}{\frac{(R - r)^{2}}{R} + 4r} \)
The term \( (R - r)^{2}/R \) is never negative and is zero only for \( R = r \), so P is maximum at \( R = r = 2\,\Omega \), which lies inside the available range 0 to \( 5\,\Omega \).
\( \displaystyle I = \frac{\varepsilon}{R + r} = \frac{12\,\mathrm{V}}{2\,\Omega + 2\,\Omega} = 3\,\mathrm{A} \)