Q15 of 35 Page 90

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

Since resistors are connected in series. So, net resistance R=


R1_+ R2


R = 2+4 = 6 Ω


Current will be calculated using ohm’s law


I= V/R


= 6/6 = 1A


Heat dissipated will be given by: H= I2Rt


for 4 Ω resistor in 5 s


H = (1)2 × 4 × 5


H = 20 J

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