11th grade physics problem...
1 answer:
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Given Information:
Power = P = 100 Watts
Voltage = V = 220 Volts
Required Information:
a) Current = I = ?
b) Resistance = R = ?
Answer:
a) Current = I = 0.4545 A
b) Resistance = R = 484 Ω
Explanation:
According to the Ohm’s law, the power dissipated in the light bulb is given by
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.
Re-arranging the above equation for current I yields,
Therefore, 0.4545 A current is flowing through the light bulb.
According to the Ohm’s law, the voltage across the light bulb is given by
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.
Re-arranging the above equation for resistance R yields,
Therefore, the resistance of the bulb is 484 Ω
Answer:
<h2>96 Joules</h2>Explanation:
We know that efficiency is the ratio of output power by input power. i.e. Efficiency describes the quality of machine or system how good it is.
Solution,
Energy input of system = 120 J
Efficiency = 80% =
Now,
According to definition,
Efficiency =
Cross multiplication:
Calculate the product
Hope this helps...
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