The current I_{1} (in A) flowing through the | vedclass

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(B) The upper branch consists of two $1\; \Omega$ resistors in parallel,which are in series with a $2\; \Omega$ resistor.The equivalent resistance of

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$0.2$

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(B) The upper branch consists of two $1\; \Omega$ resistors in parallel,which are in series with a $2\; \Omega$ resistor.The equivalent resistance of the two $1\; \Omega$ resistors in parallel is $R_{p} = \frac{1 	imes 1}{1 + 1} = 0.5\; \Omega$.The total resistance of the upper branch is $R_{upper} = 0.5\; \Omega + 2\; \Omega = 2.5\; \Omega$.The voltage across the upper branch is $V = 1\; V$.The total current $i$ flowing through the upper branch is $i = \frac{V}{R_{upper}} = \frac{1}{2.5} = 0.4\; A$.This current $i$ splits equally between the two $1\; \Omega$ resistors in parallel.Therefore,the current $I_{1}$ flowing through one of the $1\; \Omega$ resistors is $I_{1} = \frac{i}{2} = \frac{0.4}{2} = 0.2\; A$.

The current $I_{1}$ (in $A$) flowing through the $1\; \Omega$ resistor in the following circuit is:

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