# Homework Help: Circuit Analysis current and induction

1. Dec 6, 2011

### Yaaaldi

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I'm not really sure where to start with this one. I know that in steady state the capacitor will become an open circuit and the inductor a short circuit.

I first changed the current source to a thevenin equivalent of 1.2mV. I don't know how to deal with the currents though... can anyone point me in the right direction?

2. Dec 6, 2011

### Staff: Mentor

There is no need to change to a Thevenin equivalent. There is no significant advantage.

You have 12 mA and it's divided between two parallel resistive branches. One branch is a 10 Ohm resistor, the other branch's resistance you will have to determine.

You are right about how to view C and L.

3. Dec 6, 2011

### Staff: Mentor

The current through the inductor will be whatever is going through the 30 Ohm resistor.

4. Dec 6, 2011

### Staff: Mentor

This is puzzling. The option to edit my own messages seems to have disappeared just today. Is everyone finding the same?

5. Dec 6, 2011

### Yaaaldi

So I've made an attempt at part i), but I don't know if it's correct.

The resistances on the right parallel side, are 30||15 = 10. Which in series with 5 gives 15 ohms.

Then we have 10||15 which gives 6 ohms. This means the voltage going through both branches is 6*12mA = 72mV.

This means going back to the original circuit we have 72mV through both the 30 and 15 ohm resistors. The current through the inductor is therefore 72mV/30 = 2.4mA.

Have I done anything wrong here? Whats the next step in finding the voltage across the capacitor? Is it simply the voltage across the 30 ohm resistor? 72mV? Doubt it though as it seems to be worth 9 marks.

6. Dec 6, 2011

### Staff: Mentor

Voltages are ACROSS, currents go THROUGH.
Careful here; the 72 mV will be across the entire branch which comprises the 30 || 15 pair and the 5 Ω series resistor.

You could use a current divider equation to determine the current through the branch, then use it again to find the current through the 30 Ω resistor.

7. Dec 6, 2011

### Yaaaldi

Here goes another attempt. The total resistance of the right branch is 15Ω. So we have 10||15. Using the current divider equation I got the current going through the right branch to be 4.8mA. Using the current divider equation again I got 1.6mA going through the 30Ω resistor. So the current through the inductor is 1.6mA?

How do I get my hands on the voltage across the capacitor?

8. Dec 6, 2011

### Staff: Mentor

That looks good.
It looks to me like it should be the same as the voltage across the 15 Ω resistor. What's the current through the 15 Ω resistor?

9. Dec 6, 2011

### Yaaaldi

3.2mA

Could you explain why the voltage across the capacitor should be the same as the 15Ω resistor? I can't visualise it from the diagram.

Thanks!

10. Dec 6, 2011

### Staff: Mentor

They are in parallel. For that matter, at steady state the 30 Ω resistor is also in parallel with it.

11. Dec 6, 2011

### Staff: Mentor

If you trace back the ends of two or more elements, and can show that they connect directly to a common point, and can do this for both ends of the elements, of course, then those elements are electrically in parallel (whether or not they have been drawn to appear that way).

Being in parallel means they have the same voltage across them. (It also follows that the current flowing through one does NOT then flow through the other.)

12. Dec 6, 2011

### Staff: Mentor

Note: it was you who initiated the idea that only a DC analysis is called for here. So your remark that 9 marks seems disproportionate for determining the voltage across the capacitor raises doubts that only the steady-state conditions are of interest. Would you be able to formulate the second-order differential equation for an LCR circuit, and solve this to find both the transient and steady-state solutions? Have you studied calculus?