Three resistances 2Ω, 4Ω, 5Ω are combined in series and this combination is connected to a battery of 12 V emf and negligible internal resistance. The potential drop across these resistances are
  • a)
    (5,45,4.36,2.18)V
  • b)
    (2,18,5.45,4.36)V
  • c)
    (4.36,2.18,5.45)V
  • d)
    (2.18,4.36,5.45)V
Correct answer is option 'D'. Can you explain this answer?

Related Test

Answers

Bs Academy
Feb 13, 2022

Let current in the circuit is I. Then total resistance in the circuit.
R = R1 + R2 + R3 = 2 + 4 + 5 = 11Ω
∴ 
The potential drop across 2Ω resistance

The potential drop across 4Ω resistance

The potential drop across 5Ω resistance

Hence(V1, V2, V3) = (2.18, 4.36, 5.45)V

Let current in the circuit is I. Then total resistance in the circuit.R = R1 + R2 + R3 = 2 + 4 + 5 = 11Ω∴The potential drop across 2Ω resistanceThe potential drop across 4Ω resistanceThe potential drop across 5Ω resistanceHence(V1, V2, V3) = (2.18, 4.36, 5.45)V
Let current in the circuit is I. Then total resistance in the circuit.R = R1 + R2 + R3 = 2 + 4 + 5 = 11Ω∴The potential drop across 2Ω resistanceThe potential drop across 4Ω resistanceThe potential drop across 5Ω resistanceHence(V1, V2, V3) = (2.18, 4.36, 5.45)V