Voltages across Capacitors

[coobe]

Hello ! I have a problem with finding the voltages in a given circuit:



how do i find the voltages on the capacitors ? I know i should use the Kirchhoffs Voltage law somehow... But those 2 Voltage sources confuse me (ideal sources)

many thanks in advance

 

[hameeds01]

C1 = 10v
C2 = 10v
C3 = 3.33µv ≈ 0v

 

[coobe]

how did you get those values ? if i plug this circuit into Multisim ill get total different values. maybe i just put the ground on the wrong place

but besides that, im more interested in the way ill get the voltge drops

 

[hameeds01]

both ideal power supplies have the same 10v ??

 

[FvM]

- One voltage source is unspecified in the drawing.
- The voltage across C2 (and across C3 and C1) is basically undefined. You may want to make additional assumptions
about initial conditions, otherwise all voltages are unknown.

Either part of the problem has been omitted or the homework/exercise is meaningless.

 

[coobe]

oh im so sorry ! of course both voltage sources are 10V

my bad

 

[jimmy neutron]

How about this approach?

You have two (obvious) loops to apply Kirchhoff's voltage law (KVL):
Loop 1: 10 V battery and C1 and C3
Loop 2: 10 V battery and C3 and C2.

Capacitors store + q on one plate and -q on the other. In the attached figure the Q's are +ve and are placed on that side of the capacitor which receives charge when it charges up due to the battery.

Applying KVL to loop 1 gives (notation is obvious):
10 V = V1 + V3
and to loop 2 gives (have to be careful with polarities)
10 V = V2 - V3

This is two equation in three unknowns so need another equation.

I think this extra equation is [someone check this?]
Q3 = Q1 + Q2.

Use the defining relationship between capacitance, pd across and charge stored , i.e. C = Q/V.

You can either replace each Vi by Qi in the

This means that you now have three equations in three unknowns (either V1, V2 and V3 or Q1, Q2 and Q3) so can solve. If any of the Vi's come out -ve then it simply means that your guess for how the charges are placed on each capacitor was wrong.

Cheers, Jim