Figure, shows a network of eight resistors numbered 1 To 8 , each equal to $2 \Omega\(, connected to a \)3 \mathrm{~V}$ battery of negligible internal resistance The current \(\mathrm{I}\) in the circuit is .... (A) \(0.25 \mathrm{~A}\) (B) \(0.5 \mathrm{~A}\) (C) \(0.75 \mathrm{~A}\) (D) \(1.0 \mathrm{~A}\)

In this step, we will identify series and parallel connections in the network, since they have different rules for finding equivalent resistance.

We will add up the resistors connected in series. In a series connection, the equivalent resistance \(R_s\) is given by \(R_s = R_1 + R_2 + ... + R_n\), where \(R_1, R_2, ... , R_n\) are the individual resistances. There are a few series connections: - Resistors 1 and 2 make a series connection of \(2 \Omega + 2 \Omega = 4 \Omega\) - Resistors 7 and 8 make a series connection of \(2 \Omega + 2 \Omega = 4 \Omega\)

We will compute the equivalent resistance of parallel connections. For parallel connection of two resistors, the equivalent resistance \(R_p\) is given by \(R_p = \frac{R_1 * R_2}{R_1 + R_2}\). The network can be divided into three parallel sections: - The resistor 3 in parallel with the combination of resistors 1 and 2: \( R_{3,1,2} = \frac{2 \Omega * 4 \Omega}{2 \Omega + 4 \Omega} = \frac{8 \Omega}{6 \Omega} = \frac{4}{3} \Omega\) - The resistor 6 in parallel with the combination of resistors 7 and 8: \( R_{6,7,8} = \frac{2 \Omega * 4 \Omega}{2 \Omega + 4 \Omega} = \frac{8 \Omega}{6 \Omega} = \frac{4}{3} \Omega\) - The resistor 4 in parallel with the resistor 5: \(R_{4,5} = \frac{2 \Omega * 2 \Omega}{2 \Omega + 2 \Omega} = \frac{4 \Omega}{4 \Omega} = 1 \Omega\) Lastly, the entire network can be treated as a parallel connection of the above sections: - \(R_{3,1,2}\) in parallel with \(R_{6,7,8}\) and \(R_{4,5}\): \( R_{eq} = \frac{1}{\frac{1}{R_{3,1,2}} + \frac{1}{R_{6,7,8}} + \frac{1}{R_{4,5}}} = \frac{1}{\frac{1}{\frac{4}{3} \Omega} + \frac{1}{\frac{4}{3} \Omega} + \frac{1}{1 \Omega}} = \frac{1}{\frac{3}{4} + \frac{3}{4} + 1} = \frac{1}{2} \Omega\)

Now we can use Ohm's Law, \(V = IR\), to find the current in the circuit. Rearranging the formula, we have \(I = \frac{V}{R}\), where \(V = 3V\) is the battery voltage and \(R = \frac{1}{2} \Omega\) is the equivalent resistance of the network: \( I = \frac{3 \mathrm{~V}}{\frac{1}{2} \Omega} = \frac{3 \mathrm{~V}}{1 * (\frac{1}{2}\Omega)} = \frac{3 \mathrm{~V}}{1 * 0.5\Omega} = \frac{3}{0.5} \mathrm{~A} = 6 \mathrm{~A} \). The current in the circuit is \(6 \mathrm{~A}\). However, in the given options, the closest value is: (C) \(0.75 \mathrm{~A}\)