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I've been doing an analysis on a particular circuit. Nothing too complicated simply a resistor and capicator in series with a 2v supply at 500Hz / 0 degrees.

Essentally what im trying to understand and eventually draw is the phaser diagram for the above circuit.

I have made the following calculations thus far;

Capacitance of the circuit

Reactive capacitance for the circuit

Circuit Impedance in polar form

Impedance

Supply current in polar form

supply current

Voltage drop across the capacitor and resistor in polar form

voltage drop across the resistor and capacitor

I have the following (sorry for the image being massive);

What i'm unsure about is; although on the the supply voltage source states 2v, 500Hz at 0 degrees, with the capacitor in the circuit i am under the assumption that the voltage source phase angle will be shifted as a result? I am not sure how to calculate his however

Any help, advice or suggested reading would be a great help, thanks.

enter image description here

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What i'm unsure about is; although on the the supply voltage source states 2v, 500Hz at 0 degrees, with the capacitor in the circuit i am under the assumption that the voltage source phase angle will be shifted as a result?

If the voltage source is an ideal voltage source, then its output voltage (magnitude and phase) is independent of the load. So there's nothing to calculate.

If the voltage source is not an ideal source, then you need to know its output impedance. This impedance is essentially another element in series with the source and the applied load. You'd use the voltage divider rule or just include the output impedance in your nodal analysis model to determine the phase at the output of the source.

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  • \$\begingroup\$ Thanks for you clarification, i was overcomplicating it when thinking about the phaser \$\endgroup\$ – Tony GR Mar 21 '15 at 12:50
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For a series circuit it's convenient to start with the current phasor, as it's common to all components. It's best, initially, to draw the current phasor, I at 0deg and of arbitrary length (we don't know it yet). Then the resistor voltage, Vr, will be 1000I at 0deg, and the capacitor voltage,Vc, will be 677I at -90deg (from CIVIL). Adding the two voltage phasors gives the supply voltage, V, which has magnitude = SQRT(Vr^2+Vc^2), at an angle Phi = arctan(-677/1000) = -34deg. Finally, the complete phasor diagram can be rotated anticlockwise by 34deg to get the V vector along the horizontal axis (or rotate the axes by 34deg clockwise which is easier!). The supply current phasor will then be leading the supply voltage by 34deg. The magnitude of the current phasor can be found by: I = 2/SQRT(677^2+1000^2) = 1.66mA. And the resistor and capacitor voltage magnitudes can be found by Ohm's law.

Apologies for not LaTex'ing - it's on the to-do list.

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  • \$\begingroup\$ Thanks for your answer. that's pretty much the approach i took in the end, but was unsure (over complicated it in my head) \$\endgroup\$ – Tony GR Mar 21 '15 at 12:49

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