Here's the solution for the circuit problem: Figure 16: Circuit Analysis Given values: Battery voltage ($V$) = $6.0 \, \text{V}$ Resistors: $R_1 = 2 \, \Omega$, $R_2 = 3 \, \Omega$, $R_3 = 5 \, \Omega$ Circuit Configuration: The $2 \, \Omega$ and $3 \, \Omega$ resistors are connected in parallel. This parallel combination is connected in series with the $5 \, \Omega$ resistor. (d) Current through the $3 \, \Omega$ resistor Step 1: Calculate the equivalent resistance of the parallel combination ($R_p$) of the $2 \, \Omega$ and $3 \, \Omega$ resistors. $$ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} $$ $$ \frac{1}{R_p} = \frac{1}{2 \, \Omega} + \frac{1}{3 \, \Omega} = \frac{3 + 2}{6 \, \Omega} = \frac{5}{6 \, \Omega} $$ $$ R_p = \frac{6}{5} \, \Omega = 1.2 \, \Omega $$ Step 2: Calculate the total external resistance ($R_{ext}$) of the circuit. The parallel combination ($R_p$) is in series with the $5 \, \Omega$ resistor ($R_3$). $$ R_{ext} = R_p + R_3 $$ $$ R_{ext} = 1.2 \, \Omega + 5 \, \Omega = 6.2 \, \Omega $$ Step 3: Calculate the total current ($I_{total}$) flowing from the battery. $$ I_{total} = \frac{V}{R_{ext}} $$ $$ I_{total} = \frac{6.0 \, \text{V}}{6.2 \, \Omega} = \frac{60}{62} \, \text{A} = \frac{30}{31} \, \text{A} $$ Step 4: Calculate the voltage ($V_p$) across the parallel combination (which includes the $3 \, \Omega$ resistor). $$ V_p = I_{total} \times R_p $$ $$ V_p = \frac{30}{31} \, \text{A} \times 1.2 \, \Omega = \frac{30}{31} \times \frac{6}{5} \, \text{V} = \frac{6 \times 6}{31} \, \text{V} = \frac{36}{31} \, \text{V} $$ Step 5: Calculate the current ($I_{3\Omega}$) through the $3 \, \Omega$ resistor. $$ I_{3\Omega} = \frac{V_p}{3 \, \Omega} $$ $$ I_{3\Omega} = \frac{\frac{36}{31} \, \text{V}}{3 \, \Omega} = \frac{36}{31 \times 3} \, \text{A} = \frac{12}{31} \, \text{A} $$ $$ I_{3\Omega} \approx 0.387 \, \text{A} $$ The current through the $3 \, \Omega$ resistor is $\boxed{\text{0.387 A}}$. (e) Total resistance in the circuit if each cell has an internal resistance of $0.75 \, \Omega$ Assumption: The 6.0 V battery is considered as a single unit with a total internal resistance of $0.75 \, \Omega$. (If the battery consists of multiple cells, the total internal resistance would be the sum of individual cell resistances, but the number of cells is not specified.) Step 1: Identify the total external resistance ($R_{ext}$). From part (d), $R_{ext} = 6.2 \, \Omega$. Step 2: Add the internal resistance ($r$) to the external resistance to find the total resistance ($R_{total}$). $$ R_{total} = R_{ext} + r $$ $$ R_{total} = 6.2 \, \Omega + 0.75 \, \Omega $$ $$ R_{total} = 6.95 \, \Omega $$ The total resistance in the circuit is $\boxed{\text{6.95 \Omega}}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.