Question

In an electrical circuit, two resistors of 2Ω & 4Ω have been connected in series to a 6V battery. The heat dissipated by the 4Ω resistor in 5s will be

Answer 1

R= 2+4=6Ω

in series connection, same current and different voltage

total voltage= total current × total resistance

6=I×6

I=6/6= 1A

voltage across 4Ω resistor= 1×4=4v

Therefore, heat dissipated by the 4Ω resistor in 5s= (v²/R)×t= (4²/4)× 5 = 16/4 × 5 = 20J