Two lls of emf 4 V and 5 V with internal resistances 0.5 Ω and 1 Ω are connected in series with a 5.5 Ω resistor. What i
Given: Two lls of emf 4 V and 5 V with internal resistances 0.5 Ω and 1 Ω are connected in series with a 5.5 Ω resistor. What is the current through the circuit? Formula: Equivalent emf: ε_{eq = 4 + 5 = 9 V. Substitution & Calculation: Total resistance: R_{total = 0.5 + 1 + 5.5 = 7 Ω . Current: I = fracε_{eqR_{total = 9/7 approx 1.29 A . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.
Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.