# Finding voltage in a capacitors circuit

#### gbox

Joined Dec 29, 2015
42

The voltmeter shows 1V
$$q_1=c_1v_1=60*10^{-9}C$$
$$60*10^{-9}C=60*10^{-9}V_2\rightarrow V_2=1V$$
$$60*10^{-9}C=30*10^{-9}V_3\rightarrow V_2=2V$$

$$V_1+V_2+V_3=4V$$

I need to find the voltage of the battery, I can not find a way to continue
Can I say that the voltage drop on C_4 is 4V?
Can I say that C_(1+2+3) must have the same voltage drop of 4V?

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#### shteii01

Joined Feb 19, 2010
4,644
It is not clear to me how you decided that there is 4 volts across C_4.

#### gbox

Joined Dec 29, 2015
42
I didn't, I am asking if I can say that C_4 has 4V because V_1+V_2+V_3=4V ?

#### shteii01

Joined Feb 19, 2010
4,644
I didn't, I am asking if I can say that C_4 has 4V because V_1+V_2+V_3=4V ?
Yes. The two branches are in parallel. Therefore each branch has the same voltage across it.

#### J_Rod

Joined Nov 4, 2014
109
Is the voltage supply (V on the right) 70Vdc; from line 2, $$q_1 = c_1v_1 = 60 * 70^{-9}$$ C ?

Recall the rules for combining capacitors in series and in parallel. How can you determine the total charge stored in capacitors $$C_6, \, C_4, \,$$ and $$C_5$$?

#### gbox

Joined Dec 29, 2015
42
But first I must find the equivalent of C_1 and C_2 and C_3 and to find the voltage on it?

#### gbox

Joined Dec 29, 2015
42
Is the voltage supply (V on the right) 70Vdc; from line 2, $$q_1 = c_1v_1 = 60 * 70^{-9}$$ C ?

Recall the rules for combining capacitors in series and in parallel. How can you determine the total charge stored in capacitors $$C_6, \, C_4, \,$$ and $$C_5$$?
Sorry the voltmeter shows $$1V$$

In parallel is the sum is series it is like resistors in parallel

#### WBahn

Joined Mar 31, 2012
26,398
But what is the basis for claiming that V_1 + V_2 + V_3 = 4 V ?

Your very equation appears to be

q_7 = (C_7)(V_7)

yet you have given no indication what C_7 and V_7 are.

You don't give a voltage for the battery nor for the voltage source on the right hand side.

#### J_Rod

Joined Nov 4, 2014
109
Capacitors in series have the same charge, but different voltages across them, whereas capacitors in parallel distribute charge, and have the same voltage between their terminals. All of the capacitances are written in the schematic, and you are asked to determine the voltage (across the battery).

#### WBahn

Joined Mar 31, 2012
26,398
So what information do you have that is not contained in your diagram?

Do you know what the voltage of the battery is?

Do you know what the reading on the voltmeter is?

#### gbox

Joined Dec 29, 2015
42
But what is the basis for claiming that V_1 + V_2 + V_3 = 4 V ?

Your very equation appears to be

q_7 = (C_7)(V_7)

yet you have given no indication what C_7 and V_7 are.

You don't give a voltage for the battery nor for the voltage source on the right hand side.

Sorry for not mentioning there is no $$C_7$$ it is a battery

#### gbox

Joined Dec 29, 2015
42
So what information do you have that is not contained in your diagram?

Do you know what the voltage of the battery is? no

Do you know what the reading on the voltmeter is? YES 1v

#### WBahn

Joined Mar 31, 2012
26,398
Sorry for not mentioning there is no $$C_7$$ it is a battery
Then what does your second line mean, since it has a C_7 and a V_7 in it?

#### WBahn

Joined Mar 31, 2012
26,398
If you know how much voltage appears across C_1, does that tell you how much charge is on C_1?

If you know how much charge is on C_1, does that tell you how much charge is on one or more of the other capacitors? If so, which ones and why?

#### shteii01

Joined Feb 19, 2010
4,644
If you know how much voltage appears across C_1, does that tell you how much charge is on C_1?

If you know how much charge is on C_1, does that tell you how much charge is on one or more of the other capacitors? If so, which ones and why?
You misread q1 of C1 as q7, etc.

#### WBahn

Joined Mar 31, 2012
26,398
You misread q1 of C1 as q7, etc.
Okay -- so how do they write a 7? With a cross bar?

I'd recommend that the TS change how they write a 1 so that it isn't so much like a 7.

So now that you know how much charge is on C4 and C2, where did that charge have to come from? Which capacitor does that now tell you has how much charge?

#### shteii01

Joined Feb 19, 2010
4,644
Okay -- so how do they write a 7? With a cross bar?

I'd recommend that the TS change how they write a 1 so that it isn't so much like a 7.

So now that you know how much charge is on C4 and C2, where did that charge have to come from? Which capacitor does that now tell you has how much charge?
Yeah, the writing is not very good. I hope you noticed that all I did in this thread is ask one simple question and provided one simple answer. I am not interested in deciphering someone else's writing. If the person wants help and does not want to waste time on answering 100 and 1 question, then they will write clearly so that everyone would understand WTF they wrote. Obviously OP did not think of that.

#### J_Rod

Joined Nov 4, 2014
109
Recommend T.S. to post an updated schematic showing the nodes where the connected voltmeter measured 1V and the polarity of the measurement.

#### WBahn

Joined Mar 31, 2012
26,398
That would be good, though the polarity is pretty easy to figure out (but it requires that we guess the polarity of their meter and engineering is not about guessing).

There is also an assumption being made that the TS needs to be aware of -- namely that the voltages on all the capacitors is consistent with them starting out uncharged.