Two coils have self-inductance L1 = 4 mH and L2 = 1 mH respectively. The currents in the two coils are increased at the same rate. At a certain instant of time both coils are given the same power. If I1 and I2 are the currents in the two coils at that instant of time respectively, then the value of (I1/I2) is

Question

Two coils have self-inductance L1 = 4 mH and L2 = 1 mH respectively. The currents in the two coils are increased at the same rate. At a certain instant of time both coils are given the same power. If I1 and I2 are the currents in the two coils at that instant of time respectively, then the value of (I1/I2) is (A) (1/8) (B) (1/4) (C) (1/2) (D) 1

Tardigrade Admin · Accepted Answer

|ε|=L (dI/dt) L=(|ε|/dI/ dt)=(IR/dI/ dt) (IP/ I2/dI/ dt)=(P/I (dI /dt)) As P and (dI /dt) are same for both the coils ∴ (I1/I2)=(L2/L1)=(1/4)

$\frac{1}{8}$

$\frac{1}{4}$

$\frac{1}{2}$

$1$

$∣ ε ∣ = L \frac{d I}{d t}$
$L = \frac{∣ ε ∣}{d I / d t} = \frac{I R}{d I / d t}$
$\frac{I P / I ^{2}}{d I / d t} = \frac{P}{I ( d I / d t )}$
As $P$ and $(d I / d t)$ are same for both the coils
$∴ \frac{I _{1}}{I _{2}} = \frac{L _{2}}{L _{1}} = \frac{1}{4}$

81​

41​

21​

1

∣ε∣=LdtdI​ L=dI/dt∣ε∣​=dI/dtIR​ dI/dtIP/I2​=I(dI/dt)P​ As P and (dI/dt) are same for both the coils ∴I2​I1​​=L1​L2​​=41​

$\frac{1}{8}$

$\frac{1}{4}$

$\frac{1}{2}$

$1$

$∣ ε ∣ = L \frac{d I}{d t}$
$L = \frac{∣ ε ∣}{d I / d t} = \frac{I R}{d I / d t}$
$\frac{I P / I ^{2}}{d I / d t} = \frac{P}{I ( d I / d t )}$
As $P$ and $(d I / d t)$ are same for both the coils
$∴ \frac{I _{1}}{I _{2}} = \frac{L _{2}}{L _{1}} = \frac{1}{4}$