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ICSE Class 10 Physics 2017 Paper

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Question : 18 of 65
Marks: +1, -0
An electric bulb of resistance 500 Ω500 \,\Omega, draws a current of 0.4 A0.4\ \text{A}. Calculate the power of the bulb and the potential difference at its end.
Solution:  
Given: R=500 Ω, I=0.4 A, P=R=500\,\Omega,\ I=0.4\ \text{A},\ P= ?, V=V= ?
We know,
Power, P=I2RP = I^2 R
=(0.4)2×500= (0.4)^2 \times 500
=80 W= 80\ \text{W}
Potential difference,
V=RV = R
=0.4×500= 0.4 \times 500
=200 Volt.= 200\ \text{Volt}.
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