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Sagot :
Answer: 28
Explanation:
Given
Circuit breaker current is [tex]I=20\ A[/tex]
Power of the light bulb is [tex]P=100\ W[/tex]
Voltage of the DC-circuit is [tex]V=120\ V[/tex]
If the resistance are connected in parallel, they must have same voltage i.e. 120 V
So, Resistance is given by
[tex]\Rightarrow R=\dfrac{V^2}{P}\\\\\Rightarrow R=\dfrac{120^2}{100}\\\\\Rightarrow R=144\ \Omega[/tex]
For the 20 A current and 120 V battery, net resistance is
[tex]\Rightarrow R_{net}=\dfrac{120}{20}\\\\\Rightarrow R_{net}=6\ \Omega[/tex]
Suppose there are n resistance in the circuit connected in parallel.
[tex]\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}[/tex]
Thus, there can maximum of 28 bulbs.
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