Question

A certain 20-A circuit breaker trips when the current in it equals 20 A. What is the maximum number of 100-W light bulbs you can connect in parallel in an ideal 120-V dc circuit without tripping this circuit breaker

Answer 1

Answer: 28

Explanation:

Given

Circuit breaker current is $I=20\ A$

Power of the light bulb is $P=100\ W$

Voltage of the DC-circuit is $V=120\ V$

If the resistance are connected in parallel, they must have same voltage i.e. 120 V

So, Resistance is given by

$\Rightarrow R=\dfrac{V^2}{P}\\\\\Rightarrow R=\dfrac{120^2}{100}\\\\\Rightarrow R=144\ \Omega$

For the 20 A current and 120 V battery, net resistance is

$\Rightarrow R_{net}=\dfrac{120}{20}\\\\\Rightarrow R_{net}=6\ \Omega$

Suppose there are n resistance in the circuit connected in parallel.

$\Rightarrow \dfrac{144}{n}=R_{net}\\\\\Rightarrow n=\dfrac{144}{6}\\\\\Rightarrow n=28.8\approx 28\ \text{for current to be less than 20A}$

Thus, there can maximum of 28 bulbs.