COMEDK2024Morning ShiftPhysicsCurrent ElectricityActual
A cell of emf E and internal resistance r is connected to two external resistances R₁ and R₂ and a perfect ammeter. The current in the circuit is measured in four different situations: (a) without any external resistance in the circuit. (b) with resistance R₁ only (c) with R₁ and R₂ in series combination. (d) with R₁ and R₂ in parallel combination. The currents measured in the four cases in ascending order are
Options
- Aa < d < b < c
- Bc < d < b < a
- Ca < b < d < c
- Dc < b < d < a
Correct answer
D. c < b < d < a
Step-by-step solution
The current I in a circuit with a cell of emf E and internal resistance r connected to an external resistance R_ ext is given by I = E r + R_ ext . The four cases for R_ ext are: (a) R_ ext = 0 , so I_a = E r . (b) R_ ext = R₁ , so I_b = E r + R₁ . (c) R_ ext = R₁ + R₂ , so I_c = E r + R₁ + R₂ . (d) R_ ext = R₁ R₂ R₁ + R₂ , so I_d = E r + R₁ R₂ R₁ + R₂ . Comparing the denominators: Since R₁ + R₂ > R₁ > R₁ R₂ R₁ + R₂ > 0 , the denominators follow the order r + R₁ + R₂ > r + R₁ > r + R₁ R₂ R₁ + R₂ > r . Since the cur