In the given figure R₁ = 10 Ω, R₂ = 8 Ω, R₃ = 4 Ω and R₄ = 8 Ω. Battery is ideal with emf 12V. Equivalent resistant of the circuit and current supplied by battery are respectively.

Question

G1 Physics · Accepted Answer

Image R₂, R₃ and R₄ are in parallel. Therefore: 1/R₂₃₄ = 1/8 + 1/4 + 1/8 = 1/2 So: R₂₃₄ = 2 Ω This is in series with R₁ = 10 Ω. Hence: Rₑq = 10 + 2 = 12 Ω Current supplied by the battery: I = 12/12 = 1 A