Question

A coil of inductance 5H is joined to a emf 6V through resistance 10ohm at time t=0 find emf across the coil at time t=ln√2

Answer 1

here is ur answer....

Answer 2

Emf across the coil at time t=ln√2 is 3V.

Given that,

Resistance (R) = 10Ω

Voltage (V) = 6V

Inductance (L) = 5H

Initial current I₀ = (V/R) = (6/10) A = 0.6A               [From Ohms Law]

Current at time t = ln√2  ln is denoted be I.

I = I₀ (1 - e^$(\frac{-Rt}{L})$)  , where R,L,t have their respective values.

Differentiating with respect to time, we get:

dI/dt = I₀(- e^$(\frac{-Rt}{L})$ × d/dt (-Rt/L))

        = I₀(- e^$(\frac{-Rt}{L})$ × (-R/L))

        = (R/L) I₀ [e^$(\frac{-Rt}{L})$]

Putting the values of R,L,t, we get:

dI/dt = (2 × 0.6)[e^(-10ln√2/5)]

        = 1.2 [e^(-2ln√2)

        = 1.2 × 0.5   = 0.6

Emf across the coil is expressed as LdI/dt

Putting the values of L and dI/dt, we get emf as 5 × 0.6 = 3V