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ruslelena [56]
3 years ago
10

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
Physics
1 answer:
soldier1979 [14.2K]3 years ago
3 0

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.

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