# Alternating current circuits

1. Jun 20, 2004

### Dita

Two connections allow contact with two circuit elements in series inside a box, but it is not known whether the circuit elements are R,L, or C. In an attempt to find what is inside the box, you make some measurements with the following results. When a 3.0-V dc power supply is connected across the terminals , there is a direct current of 300 mA in the circuit. When a 3.0-V, 60-Hz source is connected , the current becomes 200 mA.

a) What are the two elements in the box?
b) What are their values of R, L or C?

2. Jun 21, 2004

### wisky40

the answers are one inductor and one resistor, and thier values are ~29.65mH and 10 ohms.

3. May 23, 2010

### Smok

Can anyone offer some insight on how you would go about solving this? I'm so lost in trying to solve this problem.

4. May 23, 2010

### xcvxcvvc

the direct current refers to stable conditions after the transient behavior of the circuit. First off, in DC circuits, capacitors behave like open circuits and inductors behave like short circuits after the transient response.

Therefore, the DC current represents solely the resistive element(s) inside the box, and we know that there is not a capacitor inside the box (else the current would be zero).

Using the ohm's law with the applied voltage and drawn current, we can calculate the total resistance inside the box.

After that calculation, we must find what inductive impedance (+j) must be added to the calculated resistor to draw a sinusoidal current with a peak of 200mA when a sinusoidal voltage with a peak of 3 volts is applied.

Finally, we apply the equation that calculates impedance based on frequency and inductance to find inductance.

5. May 24, 2010

### Smok

Thank you very much for the help. I'm absolutely terrible when it comes to inductance and anything related to alternating current =/

6. May 24, 2010

### Smok

So I understand everything up to this point but I keep getting stuck on calculating the L. I'm getting .0132H for some reason. I'm guessing I'm doing something horribly wrong but I'm setting Vmax as 3V, VR,max as (.2 A * 10 Ohms) and VL as (2*$$\pi$$)*60L) and plugging all that into Vmax = $$\sqrt{Vr^2+VL^2}$$

I know this is wrong but I don't know why and it's driving me nuts....

Can you or anyone else offer some insight? I hate asking for step by step instructions but my back's against the wall.

7. May 24, 2010

### xcvxcvvc

$$|Z|=\sqrt{R^2 +X^2}$$
$$X=\omega L$$
$$|I|=\frac{|V|}{|Z|}$$

8. May 25, 2010

### Smok

Ok, I was using the correct equations, it was just a stupid mathematical error. Thanks a lot for all the help.