# Step-by-Step Solution for the Circuit Analysis Let's break down the problem systematically: ## **Given:** You are asked to determine: 1. **Currents** \( I_1 \) to \( I_4 \) 2. **Voltages** \( U_1 \) to \( U_4 \) 3. **Power delivered by the current sources** ## **Circuit Summary:** - The circuit contains four resistors and three current sources. - Each resistor has a voltage and current marked. - The current sources are **2A, 3A, 5A**. --- ## **Step 1: Assigning Currents** ### **Label the Branch Currents** - \( I_1 \) through the 3Ω resistor (from top node down) - \( I_2 \) through the 1Ω resistor (left to right) - \( I_3 \) through the 4Ω resistor (bottom left to right) - \( I_4 \) through the 4Ω resistor (right, top to bottom) ### **Current Source Directions:** - **5A** enters the top node (from above) - **3A** enters from the left - **2A** enters from the bottom left --- ## **Step 2: Apply Kirchhoff's Current Law (KCL)** ### **At the Top Node:** Sum of currents entering = sum of currents leaving \[ I_1 + I_2 = 5A \] ### **At the Left Node:** \[ I_2 = 3A \] ### **At the Bottom Node:** \[ I_3 = 2A \] ### **At the Right Node:** Current entering the node is \( I_4 \), and leaving is from the 4Ω resistor. --- ## **Step 3: Apply KCL to Solve for Currents** From above: - \( I_2 = 3A \) - \( I_3 = 2A \) From top node: \[ I_1 + I_2 = 5A \implies I_1 = 5A - I_2 = 5A - 3A = 2A \] At the bottom right, the current coming out must match the sum: - Currents entering node (bottom right): \( I_3 \) and \( I_4 \) - Current leaving: \( I_1 \) (since it's a closed mesh) But let's check with loop/mesh analysis for confirmation. --- ## **Step 4: Kirchhoff's Voltage Law (KVL)** Let's write KVL for the main loops. ### **Loop 1 (top, left, bottom):** - \( U_2 \) (1Ω), \( U_1 \) (3Ω), \( U_3 \) (4Ω) ### **Voltages Across Resistors:** - \( U_1 = 3\Omega \cdot I_1 = 3 \cdot 2 = 6V \) - \( U_2 = 1\Omega \cdot I_2 = 1 \cdot 3 = 3V \) - \( U_3 = 4\Omega \cdot I_3 = 4 \cdot 2 = 8V \) - \( U_4 = 4\Omega \cdot I_4 \) (need \( I_4 \)) ### **Find \( I_4 \):** At the right node: - Entering: \( I_4 \) - Leaving: \( I_1 \) - Entering from bottom: \( I_3 \) If we consider the rightmost mesh: \[ I_1 + I_3 = I_4 \] \[ I_4 = I_1 + I_3 = 2A + 2A = 4A \] ### **Calculate \( U_4 \):** \[ U_4 = 4\Omega \cdot I_4 = 4 \cdot 4 = 16V \] --- ## **Step 5: Summary Table** | Quantity | Value | |----------|---------------| | \( I_1 \) | \( 2A \) | | \( I_2 \) | \( 3A \) | | \( I_3 \) | \( 2A \) | | \( I_4 \) | \( 4A \) | | \( U_1 \) | \( 6V \) | | \( U_2 \) | \( 3V \) | | \( U_3 \) | \( 8V \) | | \( U_4 \) | \( 16V \) | --- ## **Step 6: Power Delivered by Current Sources** **Power delivered = Voltage across the source × Current** Calculate the voltage across each current source using KVL: 1. **For 5A source (top):** - Voltage across the top mesh (sum of voltage drops): - \( U_1 \) (down), \( U_4 \) (down right), so \( V_{5A} = U_1 + U_4 = 6V + 16V = 22V \) - Power: \( P_{5A} = 22V \times 5A = 110W \) 2. **For 3A source (left):** - Across 1Ω: \( U_2 = 3V \) - Power: \( P_{3A} = 3V \times 3A = 9W \) 3. **For 2A source (bottom left):** - Across 4Ω: \( U_3 = 8V \) - Power: \( P_{2A} = 8V \times 2A = 16W \) --- ## **Final Answers** ### **Currents** - \( I_1 = 2A \) - \( I_2 = 3A \) - \( I_3 = 2A \) - \( I_4 = 4A \) ### **Voltages** - \( U_1 = 6V \) - \( U_2 = 3V \) - \( U_3 = 8V \) - \( U_4 = 16V \) ### **Power Delivered** - By 5A source: **110W** - By 3A source: **9W** - By 2A source: **16W** --- If you need a detailed step-by-step on mesh or node analysis for verification, let me know!