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Question

A torch bulb is rated 2.5Vand 750mA. Calculate its power, its resistance, and the energy consumed if this bulb is lighted for 4hours. Two identical resistors, each of resistance 2Ω are connected in the torch in series and in parallel, to a battery of 12V, calculate the ratio of power consumed in two cases.


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Solution

Step 1: Given

Torch Voltage = 2.5V

Torch current = 750mA

Time = 4hours

Resistance = 2Ω

Voltage of battery = 12V

Step 2: Formula used and calculation

Power is given by

P=VI=2.5V×.75A=1.875W

Power consumed by torch is 1.875W.

Resistance is given by

R=VI=2.5V0.75A=3.33Ω

Resistance of torch is 3.33Ω.

Energy consumed is given by

E=Pt=1.875W×4×60×60s=27000J

In series combination

Req=R1+R2=2Ω+2Ω=4Ω

V=12VI=VReq=124A=3AP1=VI=12×3W=36W

In parallel combination

1Req=1R1+1R2=12+12ΩReq=1ΩV=12VI=VReq=121A=12AP2=VI=12×12W=144W

Ratio of power in series and parallel combination

P1P2=36144=14

Hence, the ratio is 1:4.


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