Question

What minimum number of 120 W lightbulbs must be connected in parallel to a single 180 V household circuit to trip a 50.0 A circuit breaker?

______ lightbulbs

Answer 1

Concepts and reason

The concepts used for solving this problem are resistors in series, Ohm’s law, and power.

Initially, use the voltage and power to find the resistance of each bulb.

Finally, use the voltage and current to find the total resistance and the number of light bulbs.

Fundamentals

Ohm’s law states that potential difference across a capacitor is directly proportional to the current through it and the proportionality constant is called resistance.

The expression for voltage is as follows:

Here, voltage is , current is , and resistance is .

The equation for power is as follows:

Here, the power is .

The equation for power is given as follows:

Rearrange the expression to get .

$R= $

Substitute for and for .

The expression for voltage is as follows:

Rearrange the expression to get .

The expression for the total resistance is as follows:

Here, is the total resistance.

Substitute for and for .

In the parallel combination of bulbs, the total resistance is equal to the ratio of the resistance of each bulb and the number of bulbs connected.

Therefore,

Here, is the number of bulbs.

Rearrange the expression to get .

Substitute for and for .

The number of bulbs connected in parallel combination is light bulbs.

Ans:

The number of bulbs connected in parallel combination is light bulbs.