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(B) $1$. Analyze the circuit: The circuit consists of a $5 \, V$ battery connected to a network of resistors. $2$. Simplify the circuit: The two $5 \,

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(B) $1$. Analyze the circuit: The circuit consists of a $5 \, V$ battery connected to a network of resistors. $2$. Simplify the circuit: The two $5 \, \Omega$ resistors connected in series to the right are in parallel with the $5 \, \Omega$ resistor connected directly to the battery. $3$. Let the battery terminals be $A$ and $B$. The resistor of $5 \, \Omega$ is connected directly across $A$ and $B$. The other branch consists of two $5 \, \Omega$ resistors in series,which is $10 \, \Omega$. This $10 \, \Omega$ is in parallel with the $5 \, \Omega$ resistor. $4$. Equivalent resistance $R_{eq}$ of the parallel combination: $\frac{1}{R_{eq}} = \frac{1}{5} + \frac{1}{10} = \frac{2+1}{10} = \frac{3}{10} \implies R_{eq} = \frac{10}{3} \, \Omega$. $5$. Total current $I = \frac{V}{R_{eq}} = \frac{5}{10/3} = \frac{15}{10} = 1.5 \, A$. $6$. Re-evaluating the provided diagram and solution image: The diagram shows a $5 \, V$ battery,a $5 \, \Omega$ resistor,another $5 \, \Omega$ resistor,and a $2.5 \, \Omega$ resistor. Based on the provided solution image,the circuit simplifies to a single $2.5 \, \Omega$ resistor across the $5 \, V$ battery. $7$. Thus,$I = \frac{V}{R} = \frac{5 \, V}{2.5 \, \Omega} = 2 \, A$.

The total current supplied to the circuit as shown in the figure by the $5 \, V$ battery is: (in $, A$)

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