Question

A .005-uf capacitor. a 500 ohm resistor, and a 1200-uf inductor are connected in series across a
12-V,40kHz source. Would the impedance be 704 ohm and the total current be 0.017A? and how would I find the total power?

Answer 1

Step 1. Calculate the reactance of the capacitor using the formula \[X _{c}= 1/(2 Pi FC)\]
F=40 10^3
C= 5 10^-3
Pi =3.14 (approx, you more places for better answer)
Andswer will be in Ohms, prefix with -j to indicate it is capacitance reactance.

Step 2. Calculate Inductive reactance using Formula\[X _{L}=2 \pi FL\]
F=40 10^3
L=1.2 10^-3 (assuming you meant 1200-uh)
Pi = Pi
Answer will be in Ohms, prefix with j to indicate inductive reactance.
Step 3 Take the absolute difference of these two results and prefix it with the j of the
proper sign to indicate the resultant reactance (capacitance/inductive)

Step 4. Take the resultant reactance squared and sum it with the resistance of 500 Ohm squared. take the square root of that sum, that is the magnitude of the impedance.

The current will equal the voltage divided by the impedance.
The power consumed would be only the I^2 R as the reactance devices do not consume power only exchange it.

Answer 2

Error please change value of C to 5 10^-9

Answer 3

Impedance is 703 Ohms (complex resistive/capacitance. Current is .017 Amps
Power is 145.7 milliwatts (I squared R)

Answer 4

This agrees pretty much with your figures.

Answer 5

thanks alot