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bobbyw
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This problem is tougher than I though and I'm stuck on a couple of things.
There are 2 inductors, a DC power source, 2 resistors, and a switch. The circuit looks just like the image I've attached, except it will have a switch on the left of the first resistor and the values are different. sorry about that.
The DC source is 12V
The inductors are 4H, and 16H, respectively (L1 and L2).
The resistors are 12Ω and 48Ω, respectively(R1 and R2).
K=1
The switch is open for a long period of time previous to t=0 (inductors fluxed), and opened at t=0.
The problem is to find Vout(t). In the image attached, Vout would be the voltage across R2.
K=M/sqrt(L1*L2). since K is given as 1, M is sqrt(L1*L2), which is 8H.
I initially set up the loops using KVL. For the left hand side of the circuit I got:
12 = (12+s4)I1 - 8sI2
and for the right:
0 = -8I1 + (48+s16)I2
from this, the system determinant can be found, and I2 can then be found.
I am stuck right now is the switch though.
for finding V(t<0), the inductors are fluxed, thus acting as short circuits. Does this mean that the mutual inductance will come into play or can be neglected? I have thought of a couple different possibilities.
V(t<0) = 0 because the inductors are not electrically connected and no current is flowing?
V(t<0) = 12 because the inductors are acting as short circuits, and the mutual inductance will appear to be a short circuit as well, making the voltage on the left and right side of the circuit the same?
V(t<0) = a voltage divider between the two resistors if the above ^ applies?
I also need to find to V(t>=0) but the questions above should help me with that.
Thanks in advance for any help.
Homework Statement
There are 2 inductors, a DC power source, 2 resistors, and a switch. The circuit looks just like the image I've attached, except it will have a switch on the left of the first resistor and the values are different. sorry about that.
The DC source is 12V
The inductors are 4H, and 16H, respectively (L1 and L2).
The resistors are 12Ω and 48Ω, respectively(R1 and R2).
K=1
The switch is open for a long period of time previous to t=0 (inductors fluxed), and opened at t=0.
The problem is to find Vout(t). In the image attached, Vout would be the voltage across R2.
Homework Equations
K=M/sqrt(L1*L2). since K is given as 1, M is sqrt(L1*L2), which is 8H.
The Attempt at a Solution
I initially set up the loops using KVL. For the left hand side of the circuit I got:
12 = (12+s4)I1 - 8sI2
and for the right:
0 = -8I1 + (48+s16)I2
from this, the system determinant can be found, and I2 can then be found.
I am stuck right now is the switch though.
for finding V(t<0), the inductors are fluxed, thus acting as short circuits. Does this mean that the mutual inductance will come into play or can be neglected? I have thought of a couple different possibilities.
V(t<0) = 0 because the inductors are not electrically connected and no current is flowing?
V(t<0) = 12 because the inductors are acting as short circuits, and the mutual inductance will appear to be a short circuit as well, making the voltage on the left and right side of the circuit the same?
V(t<0) = a voltage divider between the two resistors if the above ^ applies?
I also need to find to V(t>=0) but the questions above should help me with that.
Thanks in advance for any help.