3 batteries and 3 capacitors

[sabak22]

Homework Statement

Three capacitors C1–C3, all initially uncharged, are placed in the circuit show above. The capacitances are C1=15.2 μF, C2=9.8 μF, C3=4.4 μF, and the battery voltages are V1=20 V, V2=20 V, V3=5 V.

What is the magnitude q1 of the charge on capacitor C1 once equilibrium has been established?

What is the magnitude q2 of the charge on capacitor C2 once equilibrium has been established?

What is the magnitude q3 of the charge on capacitor C3 once equilibrium has been established?

Homework Equations

V=Q/C
I1+I3=i2

The Attempt at a Solution

I tried applrying the kirchhoff's rule, but i don't know where to go with it... i don't even know if i am write or not.
V1+Q1/C1 + Q2/C2+V2=0
V3+Q2/C2+V2+Q3/C3=0

 

[SammyS]

No figure is visible, so it's impossible to help you.

 

[sabak22]

really sorry about that. I didnt realize

 

[SammyS]

...

The Attempt at a Solution

I tried applying the Kirchhoff rule, but i don't know where to go with it... i don't even know if i am write or not.
V1+Q1/C1 + Q2/C2+V2=0
V3+Q2/C2+V2+Q3/C3=0

For the loop with V1+Q1/C1 + Q2/C2+V2=0: Assuming the voltage drops are for traversing the loop in a Clockwise sense, there is a voltage drop across the capacitors. This equation should be:

V₁ - Q₁/C₁ - Q₂/C₂ + V₂=0 .​

Similarly, the signs for the voltages across the capacitors in the other loop equation should also be changed.

Additionally, you have Q₁ + Q₃ = Q₂, because I₁ + I₃ = I₂ .

Plugin all the values, do some algebra and see what you get.

 

[asteriatic]

I would approach this problem by first identifying the variables given in the problem: three capacitors with their respective capacitances (C1, C2, C3) and three battery voltages (V1, V2, V3). I would also take note of the fact that all capacitors are initially uncharged and that equilibrium has been established.

Next, I would use the relevant equation V=Q/C to solve for the magnitude of the charge on each capacitor. Since equilibrium has been established, the voltage across each capacitor should be equal to the voltage of the battery it is connected to. Thus, we can rewrite the equations as:

V1=Q1/C1
V2=Q2/C2
V3=Q3/C3

Solving for Q1, Q2, and Q3, we get:

Q1=V1*C1=20*15.2=304 μC
Q2=V2*C2=20*9.8=196 μC
Q3=V3*C3=5*4.4=22 μC

Therefore, the magnitude of the charge on capacitor C1 is 304 μC, the magnitude of the charge on capacitor C2 is 196 μC, and the magnitude of the charge on capacitor C3 is 22 μC.

To check our answer, we can use the Kirchhoff's rule, which states that the sum of all currents entering and leaving a junction in a circuit must be equal to zero. In this case, the only junction is the one between C1 and C2, so we can write:

I1+I3=I2

Since all capacitors are initially uncharged, the initial current through the circuit is zero. Once equilibrium is established, the current through each capacitor is equal to the charge on that capacitor divided by the time it takes to reach equilibrium. Since we are only interested in the magnitudes of the charges, we can ignore the direction of the currents and simply write:

Q1/T+Q3/T=Q2/T

Since the time taken to reach equilibrium is the same for all capacitors, we can cancel it out and rewrite the equation as:

Q1+Q3=Q2

Substituting the values we calculated earlier, we get:

304 μC+22 μC=196 μC

This equation is true, so our answer is correct.

In conclusion