Study FIGURE 2 below. Using Thevenin's method, calculate the current flowing through the 15 ",u2126" resistor.

• We need to find the potential difference between points B and D. Let's assume point C is the reference ground ($0 \text{ V}$).
• The voltage at point A is $V_T = 5 \text{ V}$.
• The voltage at point B ($V_B$) can be found using the voltage divider rule for the series combination of $R_2$ and $R_3$ connected across $V_T$: $$V_B = V_T \times \frac{R_3}{R_2 + R_3} = 5 \text{ V} \times \frac{5 \Omega}{35 \Omega + 5 \Omega} = 5 \text{ V} \times \frac{5}{40} = 0.625 \text{ V}$$
• The voltage at point D ($V_D$) can be found using the voltage divider rule for the series combination of $R_1$ and $R_4$ connected across $V_T$: $$V_D = V_T \times \frac{R_4}{R_1 + R_4} = 5 \text{ V} \times \frac{10 \Omega}{30 \Omega + 10 \Omega} = 5 \text{ V} \times \frac{10}{40} = 1.25 \text{ V}$$
• The Thevenin voltage $V_{Th}$ is the potential difference between D and B: $$V_{Th} = V_D - V_B = 1.25 \text{ V} - 0.625 \text{ V} = 0.625 \text{ V}$$

• To find $R_{Th}$, we turn off the independent voltage source ($V_T = 0 \text{ V}$), which means we replace it with a short circuit. This connects points A and C together.
• Now, looking into terminals B and D:
  • From terminal B, resistors $R_2$ ($35 \Omega$) and $R_3$ ($5 \Omega$) are connected between B and the shorted A-C line. Therefore, $R_2$ and $R_3$ are in parallel: $$R_{B,eq} = R_2 || R_3 = \frac{35 \Omega \times 5 \Omega}{35 \Omega + 5 \Omega} = \frac{175}{40} \Omega = 4.375 \Omega$$
  • From terminal D, resistors $R_1$ ($30 \Omega$) and $R_4$ ($10 \Omega$) are connected between D and the shorted A-C line. Therefore, $R_1$ and $R_4$ are in parallel: $$R_{D,eq} = R_1 || R_4 = \frac{30 \Omega \times 10 \Omega}{30 \Omega + 10 \Omega} = \frac{300}{40} \Omega = 7.5 \Omega$$
• The Thevenin resistance $R_{Th}$ is the series combination of $R_{B,eq}$ and $R_{D,eq}$ because they are connected end-to-end between terminals B and D: $$R_{Th} = R_{B,eq} + R_{D,eq} = 4.375 \Omega + 7.5 \Omega = 11.875 \Omega$$

• Now, we connect the $15 \Omega$ resistor ($R_5$) to the Thevenin equivalent circuit (a voltage source $V_{Th}$ in series with $R_{Th}$).
• The current $I_{R5}$ flowing through $R_5$ is given by Ohm's Law: $$I_{R5} = \frac{V_{Th}}{R_{Th} + R_5} = \frac{0.625 \text{ V}}{11.875 \Omega + 15 \Omega} = \frac{0.625 \text{ V}}{26.875 \Omega}$$ $$I_{R5} \approx 0.0232558 \text{ A}$$

The current flowing through the $15 \Omega$ resistor is $\boxed{\text{0.0233 A}}$.