Here are the solutions for the circuit problem: 1) Current through $3 \, \Omega$: Step 1: Identify the series combination. The $3 \, \Omega$ and $5 \, \Omega$ resistors are connected in series. Their equivalent resistance is: $$R_{series} = 3 \, \Omega + 5 \, \Omega = 8 \, \Omega$$ Step 2: Determine the voltage across the series combination. The $8 \, \Omega$ series combination is in parallel with the $2 \, \Omega$ resistor, and this entire parallel arrangement is connected directly across the $6.0 \, V$ battery. Therefore, the voltage across the $8 \, \Omega$ series combination is $6.0 \, V$. Step 3: Calculate the current through the $3 \, \Omega$ resistor. Since the $3 \, \Omega$ resistor is part of the $8 \, \Omega$ series combination, the current flowing through it is the total current through that branch. Using Ohm's Law ($I = \frac{V}{R}$): $$I_{3\Omega} = \frac{6.0 \, V}{8 \, \Omega} = 0.75 \, A$$ The current through the $3 \, \Omega$ resistor is $\boxed{\text{0.75 A}}$. 2) If each cell has an internal resistance of $0.75 \, \Omega$, determine the total resistance in the circuit: Step 1: Calculate the total external resistance ($R_{ext}$). The $3 \, \Omega$ and $5 \, \Omega$ resistors are in series: $$R_{s1} = 3 \, \Omega + 5 \, \Omega = 8 \, \Omega$$ This $R_{s1}$ is in parallel with the $2 \, \Omega$ resistor. The equivalent resistance of this parallel combination is: $$R_{ext} = \frac{R_{s1} \times 2 \, \Omega}{R_{s1} + 2 \, \Omega} = \frac{8 \, \Omega \times 2 \, \Omega}{8 \, \Omega + 2 \, \Omega} = \frac{16 \, \Omega^2}{10 \, \Omega} = 1.6 \, \Omega$$ Step 2: Calculate the total internal resistance ($r_{total}$). The battery has an EMF of $6.0 \, V$. Assuming standard $1.5 \, V$ cells, the number of cells in the battery is: $$n = \frac{6.0 \, V}{1.5 \, V} = 4 \text{ cells}$$ Each cell has an internal resistance of $0.75 \, \Omega$. So, the total internal resistance for 4 cells in series is: $$r_{total} = n \times r_{cell} = 4 \times 0.75 \, \Omega = 3.0 \, \Omega$$ Step 3: Determine the total resistance in the circuit. The total resistance in the circuit is the sum of the total external resistance and the total internal resistance: $$R_{total} = R_{ext} + r_{total} = 1.6 \, \Omega + 3.0 \, \Omega = 4.6 \, \Omega$$ The total resistance in the circuit is $\boxed{\text{4.