(A) State two applications of Joule's heating in domestic electric circuit.

Question

Aicharya · Accepted Answer

(A) Electric bulb / electric iron / electric fuse / electric heater / electric Oven (Any two) (B) (a) 1 kWh = 1000 watt x 3600 second = 3.6 × 10 6 watt second = 3.6 × 10 6 Joule (J) OR (b) $\dfrac{1}{Rp}$ = $\dfrac{1}{R1}$ + $\dfrac{1}{R2}$ + $\dfrac{1}{R3} \dfrac{1}{Rp}$ = 1/2Ω + 1/4Ω + 1/6Ω ∴ Rp = 12/11 Ω = 1.09 Ω

(A) State two applications of Joule's heating in domestic electric circuit.

Solution