Let's solve the circuit step by step for the required values. ## **Step 1: Find the Current (Ammeter Reading)** ### **Given:** - \( R_1 = 10. \, \Omega \) (top) - \( R_2 = 5.00 \, \Omega \) (bottom) - \( \Delta V_1 = 6.20 \, \text{V} \) (left battery) - \( \Delta V_2 = 3.50 \, \text{V} \) (right battery) ### **Apply Kirchhoff’s Loop Rule:** Let current \( I \) flow clockwise (through the ammeter, top resistor, batteries, and bottom resistor). **Loop Equation:** \[ -\Delta V_1 + I R_1 + \Delta V_2 + I R_2 = \] **Plug in values:** \[ -6.20 + 10.I + 3.50 + 5.00I = \] \[ (10. + 5.00)I = 6.20 - 3.50 \] \[ 15.I = 2.70 \] \[ I = \frac{2.70}{15.} = .180 \, \text{A} \] ### **Explanation:** - We used Kirchhoff’s loop law to write the equation. - Solved for the total current (ammeter reading). --- ## **Step 2: Find the Voltmeter Reading** ### **Voltmeter across the bottom resistor (\( R_2 \)):** \[ V_{R_2} = I \cdot R_2 \] \[ V_{R_2} = .180 \times 5.00 = .90 \, \text{V} \] ### **Explanation:** - The voltmeter reads the voltage drop across the bottom resistor. - Use Ohm's Law (\( V = IR \)) with calculated current. --- ## **Final Answers:** - **Ideal ammeter:** \( \boxed{180\, \text{mA}} \) - **Ideal voltmeter:** \( \boxed{.90\, \text{V}} \) If you need the "What if" part (about flipping the battery), let me know!

# Circuit Analysis Solution We will solve the circuit in two steps, addressing parts (a), (b), and (c) thoroughly. ## **Step 1: Find the Current (Ammeter Reading)** ### **Given:** - \( R_1 = 10 \, \Omega \) (top resistor) - \( R_2 = 5.00 \, \Omega \) (bottom resistor) - \( \Delta V_1 = 6.20 \, \text{V} \) (left battery) - \( \Delta V_2 = 3.50 \, \text{V} \) (right battery) ### **Apply Kirchhoff’s Loop Rule:** Assuming the current \( I \) flows clockwise through the ammeter, top resistor, and both batteries, we can write the loop equation as follows: **Loop Equation:** \[ -\Delta V_1 + I R_1 + \Delta V_2 + I R_2 = 0 \] **Substituting Known Values:** \[ -6.20 + I(10) + 3.50 + I(5.00) = 0 \] \[ (10 + 5.00)I = 6.20 - 3.50 \] \[ 15I = 2.70 \] \[ I = \frac{2.70}{15} = 0.180 \, \text{A} \] ### **Explanation:** - We applied Kirchhoff’s loop law to derive the equation for the circuit. - The total current flowing through the circuit is calculated to be \( 0.180 \, \text{A} \) or \( 180 \, \text{mA} \). --- ## **Step 2: Find the Voltmeter Reading** ### **Voltmeter Across the Bottom Resistor (\( R_2 \)):** Using Ohm's Law to find the voltage drop across \( R_2 \): \[ V_{R_2} = I \cdot R_2 \] \[ V_{R_2} = 0.180 \, \text{A} \cdot 5.00 \, \Omega \] \[ V_{R_2} = 0.90 \, \text{V} \] ### **Explanation:** - The voltmeter measures the voltage drop across the bottom resistor \( R_2 \). - By using the current calculated in Step 1, we applied Ohm's Law to determine the voltage across \( R_2 \). --- ## **Final Answers:** - **Ammeter Reading:** \( \boxed{180 \, \text{mA}} \) - **Voltmeter Reading across \( R_2 \):** \( \boxed{0.90 \, \text{V}} \) If additional scenarios or modifications are needed, feel free to ask!