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  • As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of <img alt="3\Omega" src="https://entrancecorner.oncodecogs.com/gif.latex?3%5COmeg

As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of 3\Omega. 3Ω. The currents through the resistors R_{4} and R_{5} are I_{4} and I_{5} respectively. The values of I_{4} and I_{5} are :

Option: 1

\mathrm{I}_4=\frac{2}{5} \mathrm{~A}\; and \; \mathrm{I}_5=\frac{8}{5} \mathrm{~A}


Option: 2

\mathrm{I}_4=\frac{24}{5} \mathrm{~A}$ and $\mathrm{I}_5=\frac{6}{5} \mathrm{~A}


Option: 3

\mathrm{I}_4=\frac{8}{5} \mathrm{~A}$ and $\mathrm{I}_5=\frac{2}{5} \mathrm{~A}


Option: 4

\mathrm{I}_4=\frac{6}{5} \mathrm{~A}$ and $\mathrm{I}_5=\frac{24}{5} \mathrm{~A}


Answers (1)

best_answer

\mathrm{R}_{\text {eq }}=3+1+2+\frac{20 \times 5}{25}+2 \Rightarrow \mathrm{R}_{\text {eq }}=12 \Omega

Current from battery

 \begin{aligned} & \mathrm{I}=\frac{24}{12} \Rightarrow \mathrm{I}=2 \mathrm{~A} \\ & \mathrm{I}_4+\mathrm{I}_5=2 \mathrm{~A} \end{aligned}

\mathrm{I}_4(20)=\mathrm{I}_5(5) \Rightarrow \mathrm{I}_5=4 \mathrm{I}_4 \Rightarrow \mathrm{I}_4=\frac{2}{5} \mathrm{~A} \ \ ,\ \ \ \mathrm{I}_5=\frac{8}{5} \mathrm{~A}

 

Posted by

SANGALDEEP SINGH

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