Let's solve the circuit step by step using **Kirchhoff’s Laws** and provide explanations and a diagram for each step. ### **Given Data:** - \( E_1 = 3.07\,V \) - \( E_2 = 1.45\,V \) - \( R_1 = 6.47\,\Omega \) - \( R_2 = 1.67\,\Omega \) - \( R_3 = 5.65\,\Omega \) --- ### **Step 1: Draw the Circuit** #### **Diagram** ```plaintext +----R1=6.47Ω----+ | | E1=3.07V R2=1.67Ω | | [ ] [ ] | | +------R3=5.65Ω--+ | | E2=1.45V | | | +----------------+ ``` **Explanation Box:** This is a two-loop circuit. \( E_1 \) and \( R_1 \) are in the left loop, \( E_2 \) and \( R_2 \) are in the right loop, and \( R_3 \) is shared between both loops. --- ### **Step 2: Assign Loop Currents** Let: - \( I_1 \): current in the left loop (clockwise) - \( I_2 \): current in the right loop (clockwise) **Explanation Box:** Currents in the resistors: - Through \( R_1 \): \( I_1 \) - Through \( R_2 \): \( I_2 \) - Through \( R_3 \): \( I_1 - I_2 \) (the difference, as it is shared) --- ### **Step 3: Apply Kirchoff’s Loop Law (KVL) to Each Loop** #### **Loop 1 (Left Loop):** \[ E_1 - R_1 I_1 - R_3 (I_1 - I_2) = 0 \] \[ 3.07 - 6.47 I_1 - 5.65 (I_1 - I_2) = 0 \] \[ 3.07 - 6.47 I_1 - 5.65 I_1 + 5.65 I_2 = 0 \] \[ 3.07 - (6.47 + 5.65) I_1 + 5.65 I_2 = 0 \] \[ 3.07 - 12.12 I_1 + 5.65 I_2 = 0 \] #### **Loop 2 (Right Loop):** \[ E_2 - R_2 I_2 - R_3 (I_2 - I_1) = 0 \] \[ 1.45 - 1.67 I_2 - 5.65 (I_2 - I_1) = 0 \] \[ 1.45 - 1.67 I_2 - 5.65 I_2 + 5.65 I_1 = 0 \] \[ 1.45 + 5.65 I_1 - (1.67 + 5.65) I_2 = 0 \] \[ 1.45 + 5.65 I_1 - 7.32 I_2 = 0 \] **Explanation Box:** We applied KVL to both loops, setting up two equations with two unknowns (\( I_1 \) and \( I_2 \)). --- ### **Step 4: Solve the Simultaneous Equations** \[ \begin{align*} (1) &\quad 3.07 - 12.12 I_1 + 5.65 I_2 = 0 \\ (2) &\quad 1.45 + 5.65 I_1 - 7.32 I_2 = 0 \\ \end{align*} \] Multiply equation (1) by 7.32, and equation (2) by 5.65 to eliminate \( I_2 \): \[ (1) \times 7.32: \quad 3.07 \times 7.32 - 12.12 \times 7.32 I_1 + 5.65 \times 7.32 I_2 = 0 \] \[ (2) \times 5.65: \quad 1.45 \times 5.65 + 5.65 \times 5.65 I_1 - 7.32 \times 5.65 I_2 = 0 \] Add the two equations: \[ 22.48 - 88.76 I_1 + 41.35 I_2 + 8.19 + 31.92 I_1 - 41.35 I_2 = 0 \] \( 41.35 I_2 - 41.35 I_2 = 0 \), so: \[ (22.48 + 8.19) + (-88.76 + 31.92)I_1 = 0 \] \[ 30.67 - 56.84I_1 = 0 \] \[ I_1 = \frac{30.67}{56.84} = 0.54\,A \] Plug \( I_1 \) into equation (1): \[ 3.07 - 12.12(0.54) + 5.65 I_2 = 0 \] \[ 3.07 - 6.55 + 5.65 I_2 = 0 \] \[ -3.48 + 5.65 I_2 = 0 \] \[ I_2 = \frac{3.48}{5.65} = 0.616\,A \] --- ### **Step 5: Find the Power Dissipated in Each Resistor** - **For \( R_1 \):** \( P_1 = I_1^2 R_1 = (0.54)^2 \times 6.47 = 1.88\,W \) - **For \( R_2 \):** \( P_2 = I_2^2 R_2 = (0.616)^2 \times 1.67 = 0.63\,W \) - **For \( R_3 \):** \( I_3 = I_1 - I_2 = 0.54 - 0.616 = -0.076\,A \) - \( P_3 = (I_3)^2 R_3 = (0.076)^2 \times 5.65 = 0.033\,W \) **Explanation Box:** Power dissipated in a resistor is calculated using \( P = I^2 R \). For \( R_3 \), use the difference of the currents. --- ### **Step 6: Find the Power Supplied by Each Battery** - **For \( E_1 \):** \( P_{E_1} = E_1 \times I_1 = 3.07 \times 0.54 = 1.66\,W \) - **For \( E_2 \):** \( P_{E_2} = E_2 \times I_2 = 1.45 \times 0.616 = 0.89\,W \) **Explanation Box:** Power supplied by a battery is \( P = E \times I \), where \( I \) is the current leaving the positive terminal. --- ## **Summary Table** | Component | Power Dissipated/Supplied (W) | |------------|------------------------------| | \( R_1 \) | 1.88 | | \( R_2 \) | 0.63 | | \( R_3 \) | 0.033 | | Battery 1 | 1.66 | | Battery 2 | 0.89 | --- **If you need the steps written inside diagram images or step-by-step in a different format, let me know!**

Certainly! Let's clarify and solve the circuit step-by-step, considering the configuration where: - \( R_1 \) is in the left loop with \( E_1 \) - \( R_2 \) is in the right loop with \( E_2 \) - \( R_3 \) is the common resistor shared between both loops --- ### **Step 1: Draw the Circuit Diagram** **Diagram Explanation:** - \( E_1 \) and \( R_1 \) form the left loop. - \( E_2 \) and \( R_2 \) form the right loop. - \( R_3 \) connects both loops, shared between them. **Simplified Circuit Diagram:** ``` +--------- R1 ---------+ | | E1 R3 | | +--------- R2 ---------+ | | E2 | | | +---------------------+ ``` --- ### **Step 2: Assign Loop Currents** Let: - \( I_1 \): current flowing clockwise in the **left loop** (through \( E_1 \) and \( R_1 \)) - \( I_2 \): current flowing clockwise in the **right loop** (through \( E_2 \) and \( R_2 \)) Because \( R_3 \) is shared, the current through it is: \[ I_{R_3} = I_1 - I_2 \] (assuming \( I_1 \) is flowing from the left to right, and \( I_2 \) from right to left). --- ### **Step 3: Write Kirchhoff's Voltage Law (KVL) Equations** **Left loop:** Applying KVL: \[ E_1 - R_1 I_1 - R_3 (I_1 - I_2) = 0 \] **Substitute known values:** \[ 3.07 - 6.47 I_1 - 5.65 (I_1 - I_2) = 0 \] Expand: \[ 3.07 - 6.47 I_1 - 5.65 I_1 + 5.65 I_2 = 0 \] Combine like terms: \[ 3.07 - (6.47 + 5.65) I_1 + 5.65 I_2 = 0 \] \[ 3.07 - 12.12 I_1 + 5.65 I_2 = 0 \] **Right loop:** Applying KVL: \[ E_2 - R_2 I_2 - R_3 (I_2 - I_1) = 0 \] Substitute known values: \[ 1.45 - 1.67 I_2 - 5.65 (I_2 - I_1) = 0 \] Expand: \[ 1.45 - 1.67 I_2 - 5.65 I_2 + 5.65 I_1 = 0 \] Combine like terms: \[ 1.45 + 5.65 I_1 - (1.67 + 5.65) I_2 = 0 \] \[ 1.45 + 5.65 I_1 - 7.32 I_2 = 0 \] --- ### **Step 4: Write the System of Equations** \[ \boxed{ \begin{cases} -12.12 I_1 + 5.65 I_2 = -3.07 \quad (1) \\ 5.65 I_1 - 7.32 I_2 = -1.45 \quad (2) \end{cases} } \] --- ### **Step 5: Solve for \( I_1 \) and \( I_2 \)** **Method:** - Multiply equation (1) by 7.32 - Multiply equation (2) by 12.12 - Subtract to eliminate one variable **Calculations:** Equation (1) × 7.32: \[ -12.12 \times 7.32 I_1 + 5.65 \times 7.32 I_2 = -3.07 \times 7.32 \] \[ -88.75 I_1 + 41.40 I_2 = -22.50 \] Equation (2) × 12.12: \[ 5.65 \times 12.12 I_1 - 7.32 \times 12.12 I_2 = -1.45 \times 12.12 \] \[ 68.56 I_1 - 88.75 I_2 = -17.56 \] Now subtract the second from the first: \[ (-88.75 I_1 + 41.40 I_2) - (68.56 I_1 - 88.75 I_2) = -22.50 - (-17.56) \] Simplify: \[ -88.75 I_1 - 68.56 I_1 + 41.40 I_2 + 88.75 I_2 = -4.94 \] \[ (-157.31 I_1) + (130.15 I_2) = -4.94 \] --- ### **Step 6: Solve for one current** Isolate \( I_1 \): \[ 157.31 I_1 = 130.15 I_2 + 4.94 \] Express \( I_1 \): \[ I_1 = \frac{130.15 I_2 + 4.94}{157.31} \] Plug into equation (2): \[ 5.65 I_1 - 7.32 I_2 = -1.45 \] Substitute \( I_1 \): \[ 5.65 \times \frac{130.15 I_2 + 4.94}{157.31} - 7.32 I_2 = -1.45 \] Calculate numerator: \[ \frac{5.65 \times 130.15 I_2 + 5.65 \times 4.94}{157.31} - 7.32 I_2 = -1.45 \] \[ \frac{736.60 I_2 + 27.94}{157.31} - 7.32 I_2 = -1.45 \] Multiply through by 157.31: \[ 736.60 I_2 + 27.94 - 7.32 I_2 \times 157.31 = -1.45 \times 157.31 \] Calculate: \[ 736.60 I_2 + 27.94 - (7.32 \times 157.31) I_2 = -227.94 \] \[ 736.60 I_2 + 27.94 - 1151.00 I_2 = -227.94 \] Combine like terms: \[ (736.60 - 1151.00) I_2 + 27.94 = -227.94 \] \[ -414.40 I_2 = -255.88 \] Solve for \( I_2 \): \[ I_2 = \frac{-255.88}{-414.40} \approx 0.618\,A \] Now, find \( I_1 \): \[ I_1 = \frac{130.15 \times 0.618 + 4.94}{157.31} \] Calculate numerator: \[ 130.15 \times 0.618 \approx 80.41 \] \[ 80.41 + 4.94 = 85.35 \] Then: \[ I_1 = \frac{85.35}{157.31} \approx 0.542\,A \] --- ### **Final Currents:** \[ \boxed{ \begin{aligned} I_1 &\approx 0.542\,A \\ I_2 &\approx 0.618\,A \end{aligned} } \] --- ### **Step 7: Calculate Power Dissipation and Power Supplied** - **Power dissipated in \( R_1 \):** \[ P_{R_1} = I_1^2 \times R_1 = (0.542)^2 \times 6.47 \approx 1.90\,W \] - **Power dissipated in \( R_2 \):** \[ P_{R_2} = I_2^2 \times R_2 = (0.618)^2 \times 1.67 \approx 0.64\,W \] - **Current through \( R_3 \):** \[ I_{R_3} = I_1 - I_2 = 0.542 - 0.618 = -0.076\,A \] (Note: Negative indicates the actual direction is opposite to the assumed.) - **Power dissipated in \( R_3 \):** \[ P_{R_3} = (I_{R_3})^2 \times R_3 = (0.076)^2 \times 5.65 \approx 0.026\,W \] - **Power supplied by batteries:** \[ P_{E_1} = E_1 \times I_1 = 3.07 \times 0.542 \approx 1.67\,W \] \[ P_{E_2} = E_2 \times I_2 = 1.45 \times 0.618 \approx 0.90\,W \] --- ### **Summary:** | **Component** | **Current (A)** | **Power (W)** | |----------------|----------------|--------------| | \( R_1 \) | 0.542 | 1.90 | | \( R_2 \) | 0.618 | 0.64 | | \( R_3 \) | 0.076 (opposite direction) | 0.026 | | \( E_1 \) | — | 1.67 (supplied) | | \( E_2 \) | — | 0.90 (supplied) | --- **Note:** The negative current in \( R_3 \) indicates the actual current flows from the side of the current assumed to be opposite the initial direction. --- Let me know if you'd like further details or clarification!

Certainly! Let's clarify the circuit configuration based on your description: - \( R_1 \) is in the **left loop**, connected with battery \( E_1 \). - \( R_2 \) is in the **right loop**, connected with battery \( E_2 \). - \( R_3 \) is the **common resistor** shared between the two loops, connecting the two batteries. ### **Step 1: Draw the Correct Circuit Diagram** Here's a clear representation: ``` +--- E1 --- R1 ---+ | | | R3 | | +--- E2 --- R2 ---+ ``` **Explanation:** - Battery \( E_1 \) and resistor \( R_1 \) form the left loop. - Battery \( E_2 \) and resistor \( R_2 \) form the right loop. - Resistor \( R_3 \) connects the junctions between the two batteries and resistors, acting as the shared resistor. --- ### **Step 2: Assign Loop Currents** Let: - \( I_1 \): current in the **left loop** (through \( E_1 \), \( R_1 \)) - \( I_2 \): current in the **right loop** (through \( E_2 \), \( R_2 \)) The shared resistor \( R_3 \) will carry current based on the difference: \[ I_{R_3} = I_1 - I_2 \] --- ### **Step 3: Write Kirchhoff's Voltage Law (KVL) Equations** **Left Loop:** \[ E_1 - R_1 I_1 - R_3 (I_1 - I_2) = 0 \] **Right Loop:** \[ E_2 - R_2 I_2 - R_3 (I_2 - I_1) = 0 \] --- ### **Step 4: Substitute Given Values** Given: | Quantity | Value | |------------|--------| | \( E_1 \) | 3.07 V | | \( E_2 \) | 1.45 V | | \( R_1 \) | 6.47 Ω | | \( R_2 \) | 1.67 Ω | | \( R_3 \) | 5.65 Ω | --- ### **Step 5: Expand and Simplify the Equations** **Left loop:** \[ 3.07 - 6.47 I_1 - 5.65 (I_1 - I_2) = 0 \] \[ 3.07 - 6.47 I_1 - 5.65 I_1 + 5.65 I_2 = 0 \] \[ 3.07 - (6.47 + 5.65) I_1 + 5.65 I_2 = 0 \] \[ 3.07 - 12.12 I_1 + 5.65 I_2 = 0 \] **Right loop:** \[ 1.45 - 1.67 I_2 - 5.65 (I_2 - I_1) = 0 \] \[ 1.45 - 1.67 I_2 - 5.65 I_2 + 5.65 I_1 = 0 \] \[ 1.45 + 5.65 I_1 - (1.67 + 5.65) I_2 = 0 \] \[ 1.45 + 5.65 I_1 - 7.32 I_2 = 0 \] --- ### **Step 6: Write the System of Equations** \[ \boxed{ \begin{cases} -12.12 I_1 + 5.65 I_2 = -3.07 \quad (1) \\ 5.65 I_1 - 7.32 I_2 = -1.45 \quad (2) \end{cases} } \] --- ### **Step 7: Solve the System** Multiply equation (1) by 7.32: \[ -12.12 \times 7.32 I_1 + 5.65 \times 7.32 I_2 = -3.07 \times 7.32 \] \[ -88.75 I_1 + 41.40 I_2 = -22.50 \] Multiply equation (2) by 12.12: \[ 5.65 \times 12.12 I_1 - 7.32 \times 12.12 I_2 = -1.45 \times 12.12 \] \[ 68.56 I_1 - 88.75 I_2 = -17.56 \] Now, subtract the second from the first: \[ (-88.75 I_1 + 41.40 I_2) - (68.56 I_1 - 88.75 I_2) = -22.50 - (-17.56) \] Simplify: \[ -88.75 I_1 - 68.56 I_1 + 41.40 I_2 + 88.75 I_2 = -4.94 \] \[ -157.31 I_1 + 130.15 I_2 = -4.94 \] Express \( I_1 \): \[ I_1 = \frac{130.15 I_2 + 4.94}{157.31} \] Substitute into equation (2): \[ 5.65 I_1 - 7.32 I_2 = -1.45 \] \[ 5.65 \times \frac{130.15 I_2 + 4.94}{157.31} - 7.32 I_2 = -1.45 \] Multiply through by 157.31 to clear denominator: \[ 5.65 \times (130.15 I_2 + 4.94) - 7.32 \times 157.31 I_2 = -1.45 \times 157.31 \] Calculate: \[ 736.60 I_2 + 27.94 - 1151.00 I_2 = -227.94 \] Combine like terms: \[ (736.60 - 1151.00) I_2 + 27.94 = -227.94 \] \[ -414.40 I_2 = -255.88 \] \[ I_2 = \frac{-255.88}{-414.40} \approx 0.618\,A \] Now, find \( I_1 \): \[ I_1 = \frac{130.15 \times 0.618 + 4.94}{157.31} \] \[ I_1 = \frac{80.41 + 4.94}{157.31} = \frac{85.35}{157.31} \approx 0.542\,A \] --- ### **Final Currents:** \[ \boxed{ I_1 \approx 0.542\,A, \quad I_2 \approx 0.618\,A } \] --- ### **Step 8: Calculate Power and Other Quantities** - Power dissipated in \( R_1 \): \[ P_{R_1} = I_1^2 \times R_1 = (0.542)^2 \times 6.47 \approx 1.90\,W \] - Power dissipated in \( R_2 \): \[ P_{R_2} = (0.618)^2 \times 1.67 \approx 0.64\,W \] - Power dissipated in \( R_3 \): \[ I_{R_3} = I_1 - I_2 = 0.542 - 0.618 = -0.076\,A \] Since power depends on \( I^2 \): \[ P_{R_3} = (0.076)^2 \times 5.65 \approx 0.026\,W \] - Power supplied by batteries: \[ P_{E_1} = E_1 \times I_1 = 3.07 \times 0.542 \approx 1.67\,W \] \[ P_{E_2} = E_2 \times I_2 = 1.45 \times 0.618 \approx 0.90\,W \] --- ### **Summary** | Component | Current (A) | Power (W) | |-----------|--------------|------------| | \( R_1 \) | 0.542 | 1.90 | | \( R_2 \) | 0.618 | 0.64 | | \( R_3 \) | 0.076 (opposite direction) | 0.026 | | \( E_1 \) | — | 1.67 | | \( E_2 \) | — | 0.90 | --- **This setup correctly accounts for the circuit as per your description.** If you’d like, I can also generate a diagram illustrating the currents and voltages!