1) An inductor with an inductance of L and a resistance of R is connected to the

Question

1) An inductor with an inductance of L and a resistance of R is connected to the terminals of a battery with an emf of E and negligible internal resistance.

a) Find the initial rate of increase of current in the circuit.

b) Find the rate of increase of current at the instant when the current is I.

c) Find the current at a time t after the circuit is closed.

2) You have a 0.400H inductor and a 6.00F capacitor. Suppose you take the inductor and capacitor and make a series circuit with a voltage source that has a voltage amplitude of 30.0V and an angular frequency of 250rad/s.

a) What is the phase angle of the source voltage with respect to the current?

Explanation / Answer

(1) Current is increased exponentially in the LR circuit and it is given by the following equation
i(t) = Io*(1- e-(R/L)t) ampere
(a) For intial rate
di /dt = Io*( 0 - (-R/L)e-(R/L)t) =  Io*( (R/L)e-(R/L)t)
at t = 0
di/dt = Io*(R/L)
hence initial rate will be given by = Io*(R/L)
where Io = E/R
hence rate will be = (E/R)*(R/L) = E/L
(b) As we calculated above that rate is given by
di/dt = Io*( (R/L)e-(R/L)t)
for i(t) = I = Io*(1- e-(R/L)t)
I /Io = 1- e-(R/L)t
e-(R/L)t = 1 - I/Io = 1 - I /(E/R) = 1 - IR /E = (E - IR) /E
taking reciprocal of both
e(R/L)t = E /E -IR
taking log both side
(R/L)t = ln(E / E-IR)

t = (L/R)*ln(E / E-IR)
Putting the value of t in the equation of rate

di/dt = Io*( (R/L)e-(R/L)t) = di/dt = Io*( (R/L) e-(R/L)*(L/R)ln*(E/ E-IR)   
= Io*(R/L)*[(E-IR) /E] = (E/R)*(R/L)*[(E-IR) /E]
= (E/L)*[(E-IR) /E]
(c) At any time t current in the circuit is defined by
i(t) = Io*(1- e-(R/L)t)

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