Question

A heating element connected across a battery of 100 V having an internal resistance of 1 Ω draws an initial current of 10 A at room temperature 20.0°C, which settles after a few seconds to a steady value. What is the power consumed by the battery itself after the steady temperature of 320.0°C is attained? Temperature coefficient of resistance averaged over the temperature range involved is 3.70 × 10⁻⁴°C⁻¹.

Answer 1

`I = epsilon/(R_0 + r)` Where R₀ is resistance at room temperature 20°

⇒ `R_0 = epsilon/I -1`

OR `R_0 = 100/10 -1 = R_0 = 9"Ω"`

Now Final temperature is 320°C

So, R = R₀ (1 + α ΔT)

= 9 (1 + 3.7 × 10⁻⁴ × 300)

= 10 Ohm

Power Consumed by cell (P) = i²r

= `(epsilon/(R + r))^2 xx r` Watt

= `(100/11)^2`

= 82.64