Home › Karnataka PUC II › physics › current_electricity › Two cells of emfs ε₁ = 2 V and ε₂ = 1.5 V with i…
Two cells of emfs ε₁ = 2 V and ε₂ = 1.5 V with internal resistances r₁ = 1 Ω and r₂ = 2 Ω are connected in PARALLEL across an external resistor R. The equivalent emf of the parallel combination is
A0.5 V
B2.5 V
C3.5 V
DApproximately 1.83 V (using ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂))
Answer & Solution
Correct answer: D. Approximately 1.83 V (using ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂))
For parallel cells, ε_eq = (ε₁ r₂ + ε₂ r₁)/(r₁ + r₂) = (2·2 + 1.5·1)/(1+2) = (4 + 1.5)/3 = 5.5/3 ≈ 1.83 V.
Related questions
The resistance R of a metallic conductor typicallyIn the drift model of conduction, the average drift velocity v_d of electrons in a conductA Wheatstone bridge with arms P, Q, R and S is balanced whenA 12 V battery with internal resistance 2 Ω is connected across a 4 Ω resistor. The TERMINKirchhoff's loop (or voltage) rule is a statement ofKirchhoff's junction (or current) rule is a statement ofTwo identical resistors each of resistance R are connected in parallel. The equivalent resIn the relation between current density j and electric field E inside an ohmic conductor,