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Current Electricity

Question

A resistance coil develops heat of 800 cal/sec when 20 volt is applied across its ends. Find the resistance of the coil (1 cal. = 4.2 joule). ∴ 1 cal = 4.2 J
ρ = 800 cal/sec = 800 x 4.2 J/s
ρ = 800 x 4.2 Watt.

Solution

Potenial applied across the coil, V= 20 V
Heat developed across the coil, H= 800 cal/sec

P = \frac{V^{2}}{R} or R = \frac{V^{2}}{P} = \frac{(20)^{2}}{800 \times 4.2} = 0.12 Ω .

which is the required resistance across the coil.