Figure, shows a network of seven resistors number 1 to 7, each equal to $1 \Omega\( connection to a \)4 \mathrm{~V}$ battery of negligible internal resistance The current I in the circuit is.... (A) \(0.5 \mathrm{~A}\) (B) \(1.5 \mathrm{~A}\) (C) \(2.0 \mathrm{~A}\) (D) \(3.5 \mathrm{~A}\)

Look at the given circuit and identify which resistances are in parallel and which are in series. In this circuit, we can see that there are a few groups of resistors that are connected in parallel or in series.

Resistors in parallel can be combined by using the formula: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots\). Look for sets of parallel resistances in the circuit, and label them as one effective resistor. Then, combine the parallel resistances into a single equivalent resistance. Note that the resistances in parallel should have the same voltage across them. In this circuit, resistances R4, R5, and R6 are in parallel and can be combined into one effective resistance. Since they all have resistance of \(1\Omega\), we have: \(\frac{1}{R_{eq}} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1}\), which gives us \(R_{eq} = \frac{1}{3}\Omega\).

Resistors in series can be combined by simply adding their resistances: \(R_{eq} = R_1 + R_2 + \cdots\). Now, we can combine R1, R2, R3, and the equivalent resistance we calculated in step 2 (resulting from combining R4, R5, and R6) into a single equivalent resistance. R1, R3, and the combined resistance from step 2 are in series, so we have: \(R_{eq} = 1 + \frac{1}{3} + 1 = \frac{8}{3}\Omega\). Finally, R7 and R2 are in parallel, so we combine their resistances: \(\frac{1}{R_{total}} = \frac{1}{1} + \frac{3}{8}\) which gives us \(R_{total} = \frac{8}{11}\Omega\).

Now that we have the total equivalent resistance of the circuit, we can use Ohm's law to calculate the current I in the circuit. Ohm's law states that \(I = \frac{V}{R}\), where V is the voltage and R is the resistance. Substitute the given voltage (4V) and the calculated total resistance (\(\frac{8}{11}\Omega\)) into the formula: \(I = \frac{4}{\frac{8}{11}} = \frac{44}{8} = 5.5 A\). However, none of the given options (A) \(0.5A\), (B) \(1.5A\), (C) \(2.0A\), or (D) \(3.5A\) match the calculated current of \(5.5 A\). It's possible that there is an error in the given options or the exercise itself, so it would be helpful to re-check the circuit and calculations for any errors or consult with other resources for clarification.