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Question

(A) Let the total current entering the junction be $I = 1 \, A$. The circuit consists of three parallel branches.Branch $1$ has a resistance $R_1 = 60

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(A) Let the total current entering the junction be $I = 1 \, A$. The circuit consists of three parallel branches.Branch $1$ has a resistance $R_1 = 60 \, \Omega$.Branch $2$ has a resistance $R_2 = 15 \, \Omega + 5 \, \Omega = 20 \, \Omega$.Branch $3$ has a resistance $R_3 = 10 \, \Omega$.The equivalent resistance $R_{eq}$ of the middle and bottom branches (which are in parallel) is given by:$\frac{1}{R_{23}} = \frac{1}{20} + \frac{1}{10} = \frac{1+2}{20} = \frac{3}{20} \implies R_{23} = \frac{20}{3} \, \Omega$.Now, the current $i$ in the top branch $(60 \, \Omega)$ can be found using the current divider rule between the top branch and the combined equivalent resistance of the other two branches $(R_{23} = 20/3 \, \Omega)$:$i = I 	imes \left( \frac{R_{23}}{R_1 + R_{23}} ight)$$i = 1 	imes \left( \frac{20/3}{60 + 20/3} ight) = \frac{20/3}{(180+20)/3} = \frac{20}{200} = 0.1 \, A$.Thus, the magnitude of $i$ is $0.1 \, A$.

The magnitude of $i$ in ampere unit is

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