1. **Problem statement:** We have a circuit with resistors R1=1200Ω, R2=200Ω, R3=1000Ω, R4=700Ω, R5=800Ω and a voltage source U0=12V. We need to find: - The total current $I_{ges}$ in the circuit. - The currents through resistors R4 and R5. - The voltage across resistor R3. 2. **Understanding the circuit:** The circuit has a voltage source $U_0$ connected to a top rail with series resistors R2 and R4. From the nodes between these resistors, there are shunt branches to the bottom rail through R1, R3, and R5. 3. **Step 1: Calculate equivalent resistance of the shunt branches at each node.** - At the first node (after R2), R1 is connected to ground. - At the second node (after R4), R5 is connected to ground. - R3 is connected from the middle node (between R2 and R4) to ground. 4. **Calculate the equivalent resistance of the parallel branches:** The top rail has R2 and R4 in series, with shunt resistors R1, R3, and R5 connected to ground at different nodes. 5. **Calculate the equivalent resistance of the parallel branches:** - The first node after R2 has R1 to ground. - The middle node between R2 and R4 has R3 to ground. - The node after R4 has R5 to ground. 6. **Calculate the total equivalent resistance:** We can model the circuit as follows: - The top rail is R2 and R4 in series. - At the node between R2 and R4, R3 is connected to ground. - At the node after R2, R1 is connected to ground. - At the node after R4, R5 is connected to ground. 7. **Calculate the equivalent resistance of the branches:** - The branch with R1 is connected at the node after R2. - The branch with R3 is connected at the node between R2 and R4. - The branch with R5 is connected at the node after R4. 8. **Calculate the total equivalent resistance $R_{ges}$:** We use the method of converting the circuit into a single equivalent resistance. 9. **Calculate the voltage at the node between R2 and R4 (call it $V_x$):** Let $I_{ges}$ be the total current from the source. 10. **Apply Kirchhoff's Current Law (KCL) at nodes:** At node between R2 and R4: $$\frac{V_x - V_{top}}{R_2} = \frac{V_x}{R_3} + \frac{V_x - V_{bottom}}{R_4}$$ Since $V_{top} = U_0 = 12V$ and $V_{bottom} = 0V$ (ground), 11. **Calculate currents through R1 and R5:** - Current through R1: $I_{R1} = \frac{12V - V_x}{R_1}$ - Current through R5: $I_{R5} = \frac{V_x}{R_5}$ 12. **Calculate total current $I_{ges}$:** $$I_{ges} = I_{R1} + I_{R3} + I_{R5}$$ where $$I_{R3} = \frac{V_x}{R_3}$$ 13. **Solve for $V_x$:** Using KCL at node between R2 and R4: $$\frac{12 - V_x}{R_2} = \frac{V_x}{R_3} + \frac{V_x}{R_4}$$ Multiply both sides by $R_2 R_3 R_4$ to clear denominators: $$ (12 - V_x) R_3 R_4 = V_x R_2 R_4 + V_x R_2 R_3 $$ 14. **Plug in values:** $$ (12 - V_x) \times 1000 \times 700 = V_x \times 200 \times 700 + V_x \times 200 \times 1000 $$ $$ (12 - V_x) 700000 = V_x (140000 + 200000) $$ $$ 8400000 - 700000 V_x = 340000 V_x $$ $$ 8400000 = 340000 V_x + 700000 V_x = 1040000 V_x $$ $$ V_x = \frac{8400000}{1040000} = 8.0769 V $$ 15. **Calculate currents:** $$ I_{R1} = \frac{12 - 8.0769}{1200} = \frac{3.9231}{1200} = 0.00327 A = 3.27 mA $$ $$ I_{R3} = \frac{8.0769}{1000} = 0.00808 A = 8.08 mA $$ $$ I_{R5} = \frac{8.0769}{800} = 0.01010 A = 10.10 mA $$ 16. **Calculate total current:** $$ I_{ges} = I_{R1} + I_{R3} + I_{R5} = 0.00327 + 0.00808 + 0.01010 = 0.02145 A = 21.45 mA $$ 17. **Calculate voltage across R3:** $$ U_{R3} = I_{R3} \times R_3 = 0.00808 \times 1000 = 8.08 V $$ **Final answers:** - Total current $I_{ges} = 21.45$ mA - Current through $R_4$ is the current through resistor R4 in the top rail: $$ I_{R4} = \frac{V_x}{R_4} = \frac{8.0769}{700} = 0.01154 A = 11.54 mA $$ - Current through $R_5 = 10.10$ mA - Voltage across $R_3 = 8.08$ V