# Inductor with constant current source

Hii can anyone explain how to do this. Its Given Answer as 6A. What exactly happens when a constant current source in series with inductor? TIA

EDIT: Better answer after being told by Andy Aka The voltage across both inductors is equal so the equations for each inductors inductance can be set equail to eachother so that dI1/dt*L1=L2*dI2/dt; multiplying both sides by dt and dividing dI2/dI1 and L1/L2 (where L1=3 and L2=2) then dI2/di1=1.5 so the current through the 2H inductor is1.5 the 3H inductor; so 10A=I1 +1.5*I2=2.5I; I=10/2.5=4 which is the Current at T=0 through L=3H coil and 4*1.5=6A through the 2H coil and since Current is maximum whn dI is lkargest and dI/dT is maximum 10A due to the constant current then at T=0 I1=4A & I2=6A.

From the following URL: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indtra.html#c2

Solving For Vb (Voltage across the inductor given a charging current of 10 amps and a time constant of 1.2H/10Ohms=.12=tau then at tau=1 or at the time T=.12s solving for Vb gives you a back EMF of 40-Volts and solving for current through inductor giver you 6 Amps.

The website does the calculation for you. Gives good explanations too:

• So, why is $I_2$ 6 amps? – Andy aka Dec 8 '18 at 11:16
• The back emf is not 12 volts - you forgot to divide by dt (which is indeterminate but very small for a practical circuit and zero for a simple analysis using ideal components). I'd give up now if I were you. – Andy aka Dec 8 '18 at 11:31
• I appreciate your encouragement! Fixed my mistake. The question seems like it is a bit misleading because 6-Amps flows at T=.12 not T=0. I think... – Danny Sebahar Dec 8 '18 at 11:50
• No it doesn’t. You’re going down the wrong path. – Andy aka Dec 8 '18 at 12:39
• Sounds right to me but no real need to get into the whole math things if you just think a little bit first and discount the red herrings and impossibilities! – Andy aka Dec 8 '18 at 20:56

Ignoring a few things that make this circuit largely impractical like: -

• How can a switch in series with a current source interrupt current
• The infinite voltage present across both inductors at t = 0
• There will be resistive losses in the coils
• The resistor (it is a red herring)

You are left with the basic fact that current is shared betweeen the two inductors (3 H and 2 H) inversely proportional to their inductance. So the 2 H inductor takes 1.5 times the current of the 3 H inductor and this means that $$\I_2\$$ = 6 amps.

Previously I said 6.6667 amps but that was a stupid thing to say and was a brain-fart.

In the fullness of time, the voltage across the coils would drop to zero and the current share would be determined by the inductor's respective DC resistances but the question is about $$\I_2(0^+)\$$ and not what happens after several seconds or minutes or hours.

• But for an inductor I(0-)= I(0+). Here I(0-)= 0Amps. How will it change instantaneously I didn't understand! Can u elaborate it !? – dinece Dec 9 '18 at 10:30
• It changes instantly by producing an infinite voltage across the inductors and, like I said in my answer this makes the circuit impractical and this means it's just a theoretical exercise. – Andy aka Dec 9 '18 at 10:32

If a current source drives an inductor, its state variable $$\i_L\$$ is no longer independent and the current splits between the two inductors as:

$$\I_2=I_{src}\frac{L_1}{L_1+L_2}=10\times\frac{3}{2+3}=6\,\mathrm A\$$

A quick sim shows you that.

The schematic is drawn wrong for the question. The switch should be shorting out the current source and open at T=0.

Anyway the problem is solved with the current divider equation:

I2 = 10amp * 3H/(3H+2H)

The fact that they are inductors instead of resistor is irrelevant because it the relative impedance that matters.