Let's tackle the questions step by step, with explanations at each stage and the final answers clearly labeled. --- ## (A) Calculate the value of EMF₂ ### **Step 1: Apply Kirchhoff's Law to the Top Loop (Clockwise)** The top loop includes EMF₁, R₁, R₂, EMF₂, R₃, and R₄. Let’s define the currents as shown: - \( I_1 = 8.\,A \) (through R₁ and R₂, rightward) - \( I_3 = 1.5\,A \) (through R₃, downward) - \( I_4 = 1.5\,A \) (through R₄, rightward) **KVL equation for the top loop:** \[ \text{Starting at A, moving clockwise:} \] \[ -EMF_1 + I_1R_1 + I_1R_2 + EMF_2 + I_4R_4 + I_3R_3 = \] But notice from the diagram: - \( I_1 \) splits at point A; \( I_3 \) goes through R₃ downward, \( I_4 \) continues through R₄ rightward. - At node B: \( I_1 = I_3 + I_4 \) → \( 8. = 1.5 + I_4 \) → \( I_4 = 6.5\,A \) - The current through R₄ is \( I_4 = 6.5\,A \). ### **Step 2: Substitute Values and Solve for EMF₂** Let’s traverse from left to right along the top branch (R₁, R₂, EMF₂, R₄): \[ -EMF_1 + (I_1 \cdot R_1) + (I_1 \cdot R_2) + EMF_2 + (I_4 \cdot R_4) = \] Plug in the values: - \( EMF_1 = 75.\,V \) - \( R_1 = 17.\,\Omega \) - \( R_2 = 12.\,\Omega \) - \( R_4 = 18.\,\Omega \) - \( I_1 = 8.\,A \) - \( I_4 = 6.5\,A \) \[ -75. + (8. \times 17.) + (8. \times 12.) + EMF_2 + (6.5 \times 18.) = \] \[ -75. + 136. + 96. + EMF_2 + 117. = \] \[ (136. + 96. + 117.) = 349. \] \[ -75. + 349. + EMF_2 = \] \[ EMF_2 = -274.\,V \] --- ### **Final Answer (A):** \[ \boxed{EMF_2 = -274.\,V} \] (The negative sign means its polarity is opposite to the direction assumed in the loop.) --- ## (B) Calculate the value of EMF₃ ### **Step 1: Apply Kirchhoff's Law to the Bottom Loop** The bottom loop includes R₅ and EMF₃. Given: - \( I_4 = 1.5\,A \) (through R₅, rightward) - \( R_5 = .75\,\Omega \) ### **Step 2: Traverse the Bottom Branch** \[ EMF_3 = I_4 \cdot R_5 \] \[ EMF_3 = 1.5 \times .75 = 1.125\,V \] --- ### **Final Answer (B):** \[ \boxed{EMF_3 = 1.13\,V} \] (Rounded to two decimal places.) --- ## (C) Find the current through \( R_3 \) From the diagram and given data: - The current through \( R_3 \) is explicitly labeled as \( I_3 = 1.5\,A \). --- ### **Final Answer (C):** \[ \boxed{1.5\,A} \] --- ## (D) Find the potential across \( R_2 \) ### **Step 1: Use Ohm’s Law for \( R_2 \)** \[ V_{R_2} = I_1 \cdot R_2 \] Given: - \( I_1 = 8.\,A \) - \( R_2 = 12.\,\Omega \) \[ V_{R_2} = 8. \times 12. = 96.\,V \] --- ### **Final Answer (D):** \[ \boxed{96.\,V} \] --- ## (E) Find the potential difference between point B and point A (\( V_{ba} = V_b - V_a \)) ### **Step 1: Traverse from A to B (Top Branch)** Move from A to B through \( R_2 \), EMF₂, \( R_4 \): \[ V_{ba} = -(I_1 \cdot R_2) + (-EMF_2) - (I_4 \cdot R_4) \] (Each term is negative because you move against the current or against the EMF.) Plug in the values: - \( I_1 \cdot R_2 = 8. \times 12. = 96.\,V \) - \( EMF_2 = -274.\,V \) - \( I_4 \cdot R_4 = 6.5 \times 18. = 117.\,V \) \[ V_{ba} = -96. - (-274.) - 117. \] \[ V_{ba} = -96. + 274. - 117. \] \[ V_{ba} = (274. - 96. - 117.) = (274. - 213.) = 61.\,V \] --- ### **Final Answer (E):** \[ \boxed{61.\,V} \] --- ## **Summary of Answers** - (A) \( \boxed{EMF_2 = -274.\,V} \) - (B) \( \boxed{EMF_3 = 1.13\,V} \) - (C) \( \boxed{1.5\,A} \) - (D) \( \boxed{96.\,V} \) - (E) \( \boxed{61.\,V} \)