A heating element connected across a battery of 100 V having intern...

Question

A heating element connected across a battery of 100 V having internal resistance 1 $\Omega$ draws initial current of 10 A at 20°C. What is the power consumed by battery itself after steady temperature of 320°C? ($\alpha = 3.70 	imes 10^{-4}$ °C$^{-1}$)

Accepted Answer

Initial resistance: $R_0 = V/I - r = 100/10 - 1 = 9\ \Omega$

Resistance at 320°C ($\Delta T = 300$°C):
$$R = R_0(1 + \alpha \Delta T) = 9(1 + 3.70 	imes 10^{-4} 	imes 300) = 9 	imes 1.111 = 10\ \Omega$$

Steady current:
$$I = \frac{E}{R + r} = \frac{100}{10 + 1} = \frac{100}{11} \approx 9.09 	ext{ A}$$

Power consumed by battery (internal resistance $r$):
$$P = I^2 r = \left(\frac{100}{11}ight)^2 	imes 1 \approx 82.6 	ext{ W}$$

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