Statement: Two light bulbs having the watt rating of 90 watts and 45 watts are connected in series with a 230-volt source (wall socket). Which bulb will glow brighter?
Power is rated for 230-volt input.
Solution:
We can calculate the resistance by using the Ohm's law .
From Ohm's law: R = V 2 /P
For first bulb: R = (230) 2 /90 watts = 587 Ω
For first bulb: R = (230) 2 /45 watts = 1175 Ω
The overall current flowing through the circuit will be:
I = V/R = 230/1762 = 0.130 A
In present case, the actual power consumed by the lamps can be calculated by the equation, P = I 2 R
For 90 > P = (0.130) 2 * 587 Ω = 10 watt
For 45 > P = (0.130) 2 * 1175 Ω = 19.85 watt
Power is rated for 230-volt input.
Solution:
We can calculate the resistance by using the Ohm's law .
From Ohm's law: R = V 2 /P
For first bulb: R = (230) 2 /90 watts = 587 Ω
For first bulb: R = (230) 2 /45 watts = 1175 Ω
The overall current flowing through the circuit will be:
I = V/R = 230/1762 = 0.130 A
In present case, the actual power consumed by the lamps can be calculated by the equation, P = I 2 R
For 90 > P = (0.130) 2 * 587 Ω = 10 watt
For 45 > P = (0.130) 2 * 1175 Ω = 19.85 watt
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