A resistor R=2 ohm and an inductor L=4 henry are connected in a parallel through

Question

A resistor R=2 ohm and an inductor L=4 henry are connected in a parallel through a switch S to an emf. After switch S is closed at time t=0, the emf is automatically adjusted to maintain a constant current i through S. PART (a)- Find the current i2 through the inductor as a function of time. PART (b)- At what time is the current i1 through the resistor equal to the current i2 though the inductor? PART (c)- Plot both i1 and i2 vs time on the same graph (YOU DONT NEED TO DO THIS PART IF ITS TOO MUCH WORK, JUST PART A&B; WILL BE SUFFICIENT) THANK YOU!

Explanation / Answer

after the switch is closed at time t=0, the emf of the source is automatically adjusted to maintain a constant current I through S. a) Find the current through the inductor as function of time. b) At what time the current through the resistor equal to the current through the inductor
the circuit looks like this:
a constant current source with a switch in series and then splits into a resistor and inductor in parallel after the switch is closed at time t=0, the emf of the source is automatically adjusted to maintain a constant current I through S. a) Find the current through the inductor as function of time. b) At what time the current through the resistor equal to the current through the inductor
the circuit looks like this:
a constant current source with a switch in series and then splits into a resistor and inductor in parallel after the switch is closed at time t=0, the emf of the source is automatically adjusted to maintain a constant current I through S. a) Find the current through the inductor as function of time. b) At what time the current through the resistor equal to the current through the inductor
the circuit looks like this:
a constant current source with a switch in series and then splits into a resistor and inductor in parallel After the switch is closed, the initial current will only flow through the resistor, since current through an inductor cannot change instantaneously. The steady state current will only flow through the inductor due to lower impedance. To make things easier, we can change the parallel current source and resistor to their thevenin equivalent series voltage source and resistor.
Vth = I*R
Rth = R

So now the series equations will apply:
v.L(t) = (IR)exp(-tR/L) = L*di/dt
i.L(t) = I*(1-exp(-tR/L))

Resistor & inductor currents are equal when:
i.L(t) = I*(1-exp(-tR/L)) = 0.5*I
Solving for time t gives:
t = -(L/R)ln(0.5) After the switch is closed, the initial current will only flow through the resistor, since current through an inductor cannot change instantaneously. The steady state current will only flow through the inductor due to lower impedance. To make things easier, we can change the parallel current source and resistor to their thevenin equivalent series voltage source and resistor.
Vth = I*R
Rth = R

So now the series equations will apply:
v.L(t) = (IR)exp(-tR/L) = L*di/dt
i.L(t) = I*(1-exp(-tR/L))

Resistor & inductor currents are equal when:
i.L(t) = I*(1-exp(-tR/L)) = 0.5*I
Solving for time t gives:
t = -(L/R)ln(0.5) After the switch is closed, the initial current will only flow through the resistor, since current through an inductor cannot change instantaneously. The steady state current will only flow through the inductor due to lower impedance. To make things easier, we can change the parallel current source and resistor to their thevenin equivalent series voltage source and resistor.
Vth = I*R
Rth = R

So now the series equations will apply:
v.L(t) = (IR)exp(-tR/L) = L*di/dt
i.L(t) = I*(1-exp(-tR/L))

Resistor & inductor currents are equal when:
i.L(t) = I*(1-exp(-tR/L)) = 0.5*I
Solving for time t gives:
t = -(L/R)ln(0.5)

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