Question

An electric lamp of resistance 20 ohm and a counductor of resistance 4 ohm are connected to 6v battery . calculate power of lamp

Answer 1

An electric lamp, whose resistance is 20Ω and a conductor of 4Ω resistance are connected to a 6V battery Calculate A. The total resistance of the circuit = 20Ω + 4Ω =24 Ω B. The current through the circuit = current through the bulb = current through the conductor The current through the circuit = Voltage applied/ the voltage of the battery The current through the circuit = (6/24) amp The current through the circuit =0.25 amp C. The potential difference across the electric lamp and the conductor. The potential difference across the electric lamp = 0.25 ×20 = 5 volt The potential difference across the conductor = 0.25 × 4 = 1volt

Answer 2

Answer:

Explanation:

P. D of bulb = 5v (calculate before)

R of bulb =20 ohm (given)

current, I =0.25 a

So

power of bulb ,

P=V*I

= 5*0.25

= 1.25W

OR

P=(I)^2*R (I sq * R)

=(0.25)^2*20

=1.25W

OR

P=V^2/R

= 5^2/20

= 25/20

= 1.25W