Q.29 Two inductors P and Q having inductance ratio 1:2 are connected in parallel in an electric circuit. The energy stored in the inductors P and Q are in the ratio
(A)
(B) 1:2
(C) 2:1
(D) 4:1
The energy stored in parallel inductors P and Q with inductance ratio 1:2 is in the ratio 2:1.
Problem Solution
Inductors P and Q have inductances L_P : L_Q = 1:2, so let L_P = L and L_Q = 2L. In parallel connection, voltage V across both is identical.
Energy stored in an inductor is E = \frac{1}{2} L I^2, where I is the current through it.
Current divides inversely with inductance: I_P = \frac{V}{\omega L_P} and I_Q = \frac{V}{\omega L_Q} (for AC) or proportionally in steady state, yielding I_P : I_Q = L_Q : L_P = 2:1. Thus, I_P = 2k and I_Q = k for some k. Substituting gives E_P = \frac{1}{2} L (2k)^2 = 2 L k^2 and E_Q = \frac{1}{2} (2L) k^2 = L k^2, so E_P : E_Q = 2:1.
Option Analysis
| Option | Ratio | Explanation | Status |
|---|---|---|---|
| (A) | 1:4 | Incorrect, as smaller inductance stores more energy due to higher current share. | Incorrect |
| (B) | 1:2 | Incorrect; energy ratio follows L_P I_P^2 : L_Q I_Q^2 = 1 \cdot 4 : 2 \cdot 1 = 4:2 = 2:1, not 1:2. | Incorrect |
| (C) | 2:1 | Correct, matching derivation above. | ✅ Correct |
| (D) | 4:1 | Incorrect; overestimates P’s energy share beyond inverse inductance effect. | Incorrect |
SEO Article
Inductors P and Q with inductance ratio 1:2 connected in parallel store energy in the energy stored in parallel inductors ratio of 2:1, vital for electrical engineering exams.
Core Concept
Energy in an inductor follows E = \frac{1}{2} L I^2. Parallel setup equalizes voltage, splitting current inversely: higher inductance gets less current.
Step-by-Step Derivation
Assume L_P = L, L_Q = 2L. Currents: I_P : I_Q = 2:1. Energies: E_P : E_Q = L(2I)^2 : 2L(I)^2 = 4LI^2 : 2LI^2 = 2:1.
Exam Relevance
Matches CSIR NET-style MCQs on electromagnetism. Option (C) 2:1 is key; avoid confusing with series (adds inductance) or resistor parallels.
