How to Calculate the Charge on Capacitor C5?

AI Thread Summary
To calculate the charge on capacitor C5 (Q5) with a 12V battery, the relevant equation is C=Q/V. The discussion highlights the need to find the total equivalent capacitance of the circuit, particularly focusing on capacitors C2, C3, and C4. It is noted that charge adds across parallel capacitors, but the charge does not divide evenly due to differing capacitances. Specific capacitance values are provided: C1 and C5 are 6 μF, C2 is 1.6 μF, C3 is 5.1 μF, and C4 is 4.2 μF. Understanding the distribution of charge across these capacitors is essential for determining Q5 accurately.
kjlchem
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Homework Statement



What is Q5, the charge on C5? V of battery = 12 V.

Homework Equations



C=Q/V

The Attempt at a Solution



V1 + V5 + V(2,3) = Vb
V1 + V5 + V4 = Vb

I have three unknowns and 2 equations. :(
 

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kjlchem said:

Homework Statement



What is Q5, the charge on C5? V of battery = 12 V.

Homework Equations



C=Q/V

The Attempt at a Solution



V1 + V5 + V(2,3) = Vb
V1 + V5 + V4 = Vb

I have three unknowns and 2 equations. :(
Look for the total equivalent capacitance instead. Then apply your Relevant Equation to find the charge that is pushed onto that capacitance by the voltage V.
 
Great! I got that part right.

I'm confused though on how to find Q2. I know that charge adds across a parallel circuit so Q(total) = Q2 + Q4, but I don't know Q4 either.
 
kjlchem said:
Great! I got that part right.

I'm confused though on how to find Q2. I know that charge adds across a parallel circuit so Q(total) = Q2 + Q4, but I don't know Q4 either.

Looks like more algebra will be involved :wink:

You now know the total charge Q that will be put onto the equivalent capacitance comprising C2,C3, and C4. You can work out the equivalent capacitance of C2 & C3 alone (perhaps call it C23), then determine how to divide the total charge Q across the parallel C23 and C4.
 
Better algebra than calculus. :)

I know that C23 = 1.22F and that Q2 = Q3 = Q23

I don't see how C23 is related to Q2, except for using the relevant equation. Also, I don't think that Q total divides evenly between C23 and C4 because I've already tried that.
 
Do you have any capacitance values?
 
kjlchem said:
Better algebra than calculus. :)

I know that C23 = 1.22F and that Q2 = Q3 = Q23

I don't see how C23 is related to Q2, except for using the relevant equation. Also, I don't think that Q total divides evenly between C23 and C4 because I've already tried that.

The charge won't divide evenly between the two. You need to determine how the charge distributes between two parallel capacitors (same voltage across both, different capacities).
 
Yep! C1 = C5 = 6 μF, C2 = 1.6 μF, C3 = 5.1 μF, and C4 = 4.2 μF. The battery voltage is V = 12 V.

Okay, I think I get it now.
 
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