Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
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.
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.