Home › Karnataka PUC II › Physics › Current Electricity › Two cells of emfs ε₁ and ε₂ with internal resist…
Two cells of emfs ε₁ and ε₂ with internal resistances r₁ and r₂ are connected in SERIES (positive of one to negative of other). The equivalent emf and internal resistance of the combination are
Aε_eq = (ε₁ + ε₂)/2, r_eq = r₁·r₂/(r₁+r₂)
Bε_eq = ε₁ + ε₂, r_eq = r₁·r₂/(r₁+r₂)
Cε_eq = ε₁ + ε₂, r_eq = r₁ + r₂
Dε_eq = ε₁ − ε₂, r_eq = r₁ + r₂
Answer & Solution
Correct answer: C. ε_eq = ε₁ + ε₂, r_eq = r₁ + r₂
Series cells (head-to-tail) add both the emf and the internal resistance. Parallel cells would invert and combine reciprocally.
Related questions
The reactance of a capacitor $C$ at frequency $f$ in an a.c. circuit is:A bird perched on a single high voltage wire is unharmed because:The two conservation laws embodied in Kirchhoff's rules are conservation of:The resistance offered by the cell itself is called:Which device detects the current in a Wheatstone bridge?In a Wheatstone bridge, the source is connected across the:The Wheatstone bridge is made up of how many resistors?At a junction, the current entering equals the current: