Solving a Node Analysis Problem with 0.632V Voltage Source

[dwn]

Homework Statement

Attached image


Using node analysis:
ix(0-) = (v-4)/10 + v/3 + v/5 = 0
v = 0.632

ix(0-) = -v/3 = -0.211 A
I don't understand why they didn't use a current divider since there is a 5 ohm resistor in parallel with the 3 ohm. Current changes for resistors that are in parallel...

il(0-) = v/5 = -0.63/5 = 0.126A

Then when it comes to solve for v(0+) = di/dt = d(0.126)/dt = 0
How can the voltage across an inductor be zero when it's charged by the 4V source for t < 0??

 

[gneill]

Homework Statement

Attached image


Using node analysis:
ix(0-) = (v-4)/10 + v/3 + v/5 = 0
v = 0.632

ix(0-) = -v/3 = -0.211 A
I don't understand why they didn't use a current divider since there is a 5 ohm resistor in parallel with the 3 ohm. Current changes for resistors that are in parallel...

They could have, but it would be as much or more work since you need the total current to begin with (or make a Norton equivalent out of the 4V and 10 Ω resistor and have a three leg current divider). Nodal analysis is pretty straightforward and doesn't require any electronic gymnastics to apply to the circuit as given.

il(0-) = v/5 = -0.63/5 = 0.126A

Then when it comes to solve for v(0+) = di/dt = d(0.126)/dt = 0
How can the voltage across an inductor be zero when it's charged by the 4V source for t < 0??

Why do you say that di/dt = 0?

When the switch opens the inductor starts with some initial current. The inductor will produce whatever EMF is required to try to maintain that current in the first instant. But the current will decay from there (and so will the EMF) as the resistors bleed away energy (heat).

 

[dwn]

Why do you say that di/dt = 0?

When the switch opens the inductor starts with some initial current. The inductor will produce whatever EMF is required to try to maintain that current in the first instant. But the current will decay from there (and so will the EMF) as the resistors bleed away energy (heat).

Exactly, that's what I was thinking, but the solution manual says otherwise. I think they made the error and used the constant iL(0) = 0.126A instead of the function 0.126e^(-t/tau) when trying to find vL(0+). Safe assessment?

 

[gneill]

Exactly, that's what I was thinking, but the solution manual says otherwise. I think they made the error and used the constant iL(0) = 0.126A instead of the function 0.126e^(-t/tau) when trying to find vL(0+). Safe assessment?

Dunno. But it is certainly true that for the instant t = 0+ the current will be the same as the current at t = 0-. That is, it will be 0.126 A for that instant and can be treated as a constant value for the voltage calculation at t = 0+.