In: Physics
How many 65-W lightbulbs can be connected in parallel across a potential difference of 80 V before the total current in the circuit exceeds 2.0 A ?
Total current in the circuit is = I
Resistances in each bulb is = R
Power each bulb is, P = 65 W
Emf of the battery = E = 80 volt
As, the bulbs are connected in parallel across the battery, then the potential difference across each bulb is E = 80 volt
Now. power (P) loss in a resistance = voltage difference (E) across the resistance current (i) across the resistance.
So, current in each bulb is, i = P / E = (65 W) / (80 volt) = (65 / 80) Amp = 0.8125 Amp
If there are n number of bulbs, then the total current in the circuit, I = n i .
Here, value of total current (I) should be less than or equal to 2.0 Amp.
So,
But, n should be an integer.
So, we should take the value of n = 2 .
So, maximum 2 bulbs can be connected before the total current in the circuit exceeds 2.0 Amp.