# Reactive circuits, working out voltage of a voltage source

#### cakfef

Joined Jan 29, 2016
5
How do I work out the Vs ? I have achieved an answer of 120V.

#### Jony130

Joined Feb 17, 2009
5,445
From voltage divider.
Vs = 80V * (1 + (R1||R2 + R4 )/R6) = 80V *(1 +30/20) = 80V*2.5 = 200V

#### Dodgydave

Joined Jun 22, 2012
10,827
You don't know the DC resistance of the inductor L2, so you can't find it.

If its a short circuit, it will form a voltage divider with the resistors R4,6 so it will be half of the supply voltage, which is 160V.

#### chuckey

Joined Jun 4, 2007
75
C2 isolates the inductor so the steady state current is zero around this bit.
Frank

#### WBahn

Joined Mar 31, 2012
28,469
MOD NOTE: Moved to Homework Help.

#### WBahn

Joined Mar 31, 2012
28,469
You don't know the DC resistance of the inductor L2, so you can't find it.
It's probably a reasonable assumption that the DC resistance of L2 is << R6, which is 20 kΩ.

#### WBahn

Joined Mar 31, 2012
28,469
C2 isolates the inductor so the steady state current is zero around this bit.
Frank
C2 isolates L1 in DC steady state, but not L2.

#### WBahn

Joined Mar 31, 2012
28,469
If its a short circuit, it will form a voltage divider with the resistors R4,6 so it will be half of the supply voltage, which is 160V.
So R1 and R2 don't factor into it at all?

#### shteii01

Joined Feb 19, 2010
4,644
Assuming Vs is DC voltage source.

1) All capacitors are fully charged and act as open circuits.
2) All inductors are fully charged and act as short circuits.

Redraw the circuit, replace all capacitors with opens circuits, replaced all inductors with a piece of wire (short circuit).

It looks like you will have 80 volts across R6. Reduce the circuit to two resistors, R6 and R whatever. The two resistors form voltage divider. Formula for voltage divider has 4 variables, you know 3 of those variables (R6, 80 volts across R6, R whatever), solve for Vs.

#### cakfef

Joined Jan 29, 2016
5
Thankyou good people Also on the same question how do I work out the power dissipation, Ive worked it out and got 512mV but don't know if its correct. Thanks

#### shteii01

Joined Feb 19, 2010
4,644
Thankyou good people Also on the same question how do I work out the power dissipation, Ive worked it out and got 512mV but don't know if its correct. Thanks
Power is measured in Watt. Why is your result in volts?

Power here is Voltage times Current.
P=VI
Since you don't know any of the currents, but you do know voltages and resistances, you can do a bit of manipulation:
V=IR
I=V/R
Therefore Power can be found by:
P=VI=V(V/R)=$$\frac{V^{2}}{R}$$

From the simplified circuit (the one with just two resistors, R6 and R whatever) work your way back to the full circuit with all the resistors. Find the voltages across each resistor and use the resistor values that were given to you in the problem description to find power dissipation of each resistor.

For example:
You already know that R6 is 20 kOhm, votage across R6 is 80 V.
So Power dissipated by R6 is then=80^2/20000=0.32 W

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