RMS Current in Circuit with Capacitor, Inductor, Resistor, AC Source

[thenewbosco]

Four circuit elements—a capacitor, an inductor, a resistor, and an AC source—are connected together in various ways.
First the capacitor is connected to the source, and the rms current is found to be 25.1 mA. The capacitor is disconnected and discharged, and then connected in series with the resistor and the source, making the rms current 15.7 mA. The circuit is disconnected and the capacitor discharged. The capacitor is then connected in series with the inductor and the source, making the rms current 68.2 mA. After the circuit is disconnected and the capacitor discharged, all four circuit elements are connected together in a series loop.
What is the rms current in the circuit?

i was thinking to use the following: $$I_{rms}=\frac{\Delta V_{rms}}{Z}$$ where Z is the impedance given by $$Z=\sqrt{R^2 + (X_L-X_C)^2$$, however i don't have the root mean square voltage...

any hints on how to approach this problem... thanks

 

[plusaf]

just a shot in the dark... it's late here...
start with an assumption of a unity driving signal... one volt rms and one Hz. i'd bet that in the end the frequency and voltage drop out of the final solution.

 

[kyle8921]

I would approach this problem by first analyzing the individual circuits separately and then combining them to find the overall rms current.

In the first circuit, where the capacitor is connected directly to the source, the rms current is 25.1 mA. This is because when a capacitor is connected to an AC source, it allows the flow of alternating current with a certain magnitude depending on the capacitance value. The presence of an inductor or resistor does not affect the current in this circuit.

In the second circuit, where the capacitor is connected in series with the resistor and source, the rms current decreases to 15.7 mA. This is because the resistor limits the flow of current in the circuit, resulting in a lower rms current. The presence of the capacitor also affects the current as it resists changes in voltage, causing a phase shift.

Similarly, in the third circuit, where the capacitor is connected in series with the inductor and source, the rms current increases to 68.2 mA. This is because the inductor resists changes in current, causing a phase shift and increasing the overall current.

To find the overall rms current in the final circuit where all four elements are connected in series, we need to consider the impedance of the circuit. The impedance is given by Z = √(R^2 + (XL - XC)^2). Here, R is the resistance of the resistor, XL is the reactance of the inductor (XL = 2πfL), and XC is the reactance of the capacitor (XC = 1/2πfC). We can use this impedance value to calculate the overall rms current using the formula I_{rms} = ΔV_{rms}/Z, where ΔV_{rms} is the root mean square voltage of the AC source.

To find the root mean square voltage, we can use the fact that the overall power in the circuit remains constant. Therefore, we can equate the power in the individual circuits to find the root mean square voltage. This can be done by using the formula P = I^2R (for the resistor circuit), P = I^2(XL - XC) (for the inductor and capacitor circuit), and P = I^2XC (for the capacitor and source circuit). By equating these powers, we can find the value of ΔV_{rms} and then use it in the formula to find the overall rms current in the final circuit