Use the superposition method to calculate the current flowing through the 25 Ω resistor shown in FIGURE 1 below:

• The $25 \Omega$ and $21 \Omega$ resistors are in parallel: $$R_{p1} = \frac{25 \Omega \times 21 \Omega}{25 \Omega + 21 \Omega} = \frac{525}{46} \Omega \approx 11.413 \Omega$$
• This $R_{p1}$ is in series with the $22 \Omega$ resistor: $$R_{s1} = 22 \Omega + R_{p1} = 22 \Omega + \frac{525}{46} \Omega = \frac{1012 + 525}{46} \Omega = \frac{1537}{46} \Omega \approx 33.413 \Omega$$
• This $R_{s1}$ is in parallel with the $30 \Omega$ resistor: $$R_{p2} = \frac{30 \Omega \times R_{s1}}{30 \Omega + R_{s1}} = \frac{30 \Omega \times \frac{1537}{46} \Omega}{30 \Omega + \frac{1537}{46} \Omega} = \frac{46110}{1380 + 1537} \Omega = \frac{46110}{2917} \Omega \approx 15.807 \Omega$$
• The total resistance seen by $V_1$ is $60 \Omega$ in series with $R_{p2}$: $$R_{total1} = 60 \Omega + R_{p2} = 60 \Omega + \frac{46110}{2917} \Omega = \frac{175020 + 46110}{2917} \Omega = \frac{221130}{2917} \Omega \approx 75.807 \Omega$$
• The total current from $V_1$ is: $$I_{total1} = \frac{V_1}{R_{total1}} = \frac{11 \text{ V}}{\frac{221130}{2917} \Omega} = \frac{32087}{221130} \text{ A} \approx 0.1451 \text{ A}$$
• Using current division, the current flowing through the branch containing $R_{s1}$ (which includes the $25 \Omega$ resistor) is: $$I_{branch1} = I_{total1} \times \frac{30 \Omega}{30 \Omega + R_{s1}} = \frac{32087}{221130} \text{ A} \times \frac{30 \Omega}{30 \Omega + \frac{1537}{46} \Omega} = \frac{32087}{221130} \times \frac{1380}{2917} \text{ A} \approx 0.0686 \text{ A}$$
• This current $I_{branch1}$ then splits between $25 \Omega$ and $21 \Omega$. The current through the $25 \Omega$ resistor is: $$I_{25,V1} = I_{branch1} \times \frac{21 \Omega}{25 \Omega + 21 \Omega} = \frac{44279460}{645065010} \text{ A} \times \frac{21}{46} \approx 0.03133 \text{ A}$$ The direction of this current is downwards.

• The $60 \Omega$ and $30 \Omega$ resistors are in parallel: $$R_{p3} = \frac{60 \Omega \times 30 \Omega}{60 \Omega + 30 \Omega} = \frac{1800}{90} \Omega = 20 \Omega$$
• This $R_{p3}$ is in series with the $22 \Omega$ resistor: $$R_{s2} = 22 \Omega + R_{p3} = 22 \Omega + 20 \Omega = 42 \Omega$$
• This $R_{s2}$ is in parallel with the $25 \Omega$ resistor: $$R_{p4} = \frac{25 \Omega \times R_{s2}}{25 \Omega + R_{s2}} = \frac{25 \Omega \times 42 \Omega}{25 \Omega + 42 \Omega} = \frac{1050}{67} \Omega \approx 15.672 \Omega$$
• The total resistance seen by $V_2$ is $21 \Omega$ in series with $R_{p4}$: $$R_{total2} = 21 \Omega + R_{p4} = 21 \Omega + \frac{1050}{67} \Omega = \frac{1407 + 1050}{67} \Omega = \frac{2457}{67} \Omega \approx 36.672 \Omega$$
• The total current from $V_2$ is: $$I_{total2} = \frac{V_2}{R_{total2}} = \frac{9 \text{ V}}{\frac{2457}{67} \Omega} = \frac{603}{2457} \text{ A} \approx 0.2454 \text{ A}$$
• Using current division, the current flowing through the $25 \Omega$ resistor is: $$I_{25,V2} = I_{total2} \times \frac{R_{s2}}{25 \Omega + R_{s2}} = \frac{603}{2457} \text{ A} \times \frac{42 \Omega}{25 \Omega + 42 \Omega} = \frac{603}{2457} \times \frac{42}{67} \text{ A} \approx 0.15385 \text{ A}$$ The direction of this current is downwards.

Both currents $I_{25,V1}$ and $I_{25,V2}$ flow in the same direction (downwards) through the $25 \Omega$ resistor.

The current flowing through the $25 \Omega$ resistor is $\boxed{\text{0.1852 A}}$.

• Assume the bottom wire (connected to C and D) is ground ($0 \text{ V}$). Then $V_A = 5 \text{ V}$.
• For the path A-B-C, the voltage at B is found using the voltage divider rule: $$V_B = V_A \times \frac{R_3}{R_2 + R_3} = 5 \text{ V} \times \frac{5 \Omega}{35 \Omega + 5 \Omega} = 5 \text{ V} \times \frac{5}{40} = 0.625 \text{ V}$$
• For the path A-D-C, the voltage at D is found using the voltage divider rule: $$V_D = V_A \times \frac{R_4}{R_1 + R_4} = 5 \text{ V} \times \frac{10 \Omega}{30 \Omega + 10 \Omega} = 5 \text{ V} \times \frac{10}{40} = 1.25 \text{ V}$$
• The Thevenin voltage $V_{Th}$ is the potential difference between B and D. Let's define it as $V_{DB}$ (since $V_D > V_B$). $$V_{Th} = V_{DB} = V_D - V_B = 1.25 \text{ V} - 0.625 \text{ V} = 0.625 \text{ V}$$

• Short-circuit the voltage source $V_T$. This means points A and C are now electrically connected.
• Looking from terminal B, $R_2$ is connected to A and $R_3$ is connected to C. Since A and C are shorted, $R_2$ and $R_3$ are in parallel: $$R_{BC'} = R_2 || R_3 = \frac{35 \Omega \times 5 \Omega}{35 \Omega + 5 \Omega} = \frac{175}{40} \Omega = 4.375 \Omega$$
• Looking from terminal D, $R_1$ is connected to A and $R_4$ is connected to C. Since A and C are shorted, $R_1$ and $R_4$ are in parallel: $$R_{DC'} = R_1 || R_4 = \frac{30 \Omega \times 10 \Omega}{30 \Omega + 10 \Omega} = \frac{300}{40} \Omega = 7.5 \Omega$$
• The Thevenin resistance $R_{Th}$ is the sum of these two parallel combinations, as they are effectively in series between B and D: $$R_{Th} = R_{BC'} + R_{DC'} = 4.375 \Omega + 7.5 \Omega = 11.875 \Omega$$

• Now, connect the $15 \Omega$ resistor ($R_5$) to the Thevenin equivalent circuit. The current $I_{R5}$ is given by: $$I_{R5} = \frac{V_{Th}}{R_{Th} + R_5} = \frac{0.625 \text{ V}}{11.875 \Omega + 15 \Omega} = \frac{0.625 \text{ V}}{26.875 \Omega} \approx 0.0232558 \text{ A}$$

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