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(A) The circuit can be simplified by identifying the series and parallel combinations of resistors.First,the $2 \Omega$ and $4 \Omega$ resistors in th

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(A) The circuit can be simplified by identifying the series and parallel combinations of resistors.First,the $2 \Omega$ and $4 \Omega$ resistors in the right branch are in series,giving $2+4=6 \Omega$.This $6 \Omega$ is in parallel with the $6 \Omega$ resistor in the middle branch,giving an equivalent resistance of $\frac{6 	imes 6}{6+6} = 3 \Omega$.This $3 \Omega$ is in series with the $12 \Omega$ resistor,but looking at the simplified circuit diagram provided,the total resistance of the network connected to the $2 \Omega$ resistor is $R_{eq} = 2 + (6 || 6) = 2 + 3 = 5 \Omega$.Wait,re-evaluating the circuit: The total resistance $R_{total} = 2 \Omega + ( (6+12) || (2+4) ) = 2 + (18 || 6) = 2 + \frac{18 	imes 6}{18+6} = 2 + \frac{108}{24} = 2 + 4.5 = 6.5 \Omega$.Using Ohm's law,$I = \frac{V}{R_{total}} = \frac{6.5 	ext{ V}}{6.5 \Omega} = 1 	ext{ A}$.

In the following circuit,the current through the resistor $R(=2 \Omega)$ is $I$ Amperes. The value of $I$ is

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