Parallel and Series circuits w/ multiple batteries

Gothy

Hello, this is my first time posting on this forum and I have seeked help on this forum before on previous assignments for my physics class.

There are four problems that I do not understand clearly or do not understand how the image is portrayed in a blueprint format

Problem 1
1. The problem statement, all variables and given/known data
http://img291.imageshack.us/img291/4382/problem39.jpg
Find i

2. Relevant equations
Kirchhoff's rules


3. The attempt at a solution
I can find 2 loops with Kirchhoff's rules, but it is not enough to find i.
6 - 3.5i₂ - 2.2i = 0; 6 - 3.5i₁ - 2.2i = 0; I understand I need to find a loop without i so I can eliminate i₂ or i₁ from one of the equations, but I do not understand the concept of two batteries facing each other.
I tried using
i₂=i₃+i or i = i₂ - i₃ and i just comes out to 0
Problem 2
1. The problem statement, all variables and given/known data
http://img175.imageshack.us/i/problem30.jpg/

If the emf of the battery is 15 V, and each resistance is 2, what is the power consumed by bulb B?

Bulb D is then removed from its socket.How does the brightness of bulb A change?

How does the brightness of bulb B change when bulb D is removed from its socket?
2. Relevant equations
not sure?

3. The attempt at a solution
The picture itself is confusing to me. I do not get how to put the light bulb picture into a blueprint format. Any information on simplifying this image would be amazing.

Problem 3
1. The problem statement, all variables and given/known data
http://img687.imageshack.us/i/problem25.jpg/
Find the equivalent resistance R between points A and B of the resistor network.
2. Relevant equations
series circuit resistance = R₁+R₂+R₃....
Parallel circuit resistance= [(1/R₁) + (1/R₂) + (1/R₃)]^-1
3. The attempt at a solution
(1/(41+21+26)+1/33)^-1= 264/11
[(1/(21+ 264/11)) + 1/36]^-1 = 1980/59 = 33.55932203

Solved Final answer = 20


Problem 4
1. The problem statement, all variables and given/known data
http://img85.imageshack.us/i/problem22.jpg/

R = [(675-486)/5.8] - .36 - .16 - 1.15= 30.9162069

Find the potential difference V_xy = V_x − V_y between points X and Y

How much energy UE is dissipated by the 1.15 in 55 s ?
2. Relevant equations
V_xy = V_x − V_y
P=IV
V=IR

3. The attempt at a solution
Once again the negative sides facing each other completely throw me off and I have no Idea what to do.

vela

There's nothing particularly special about having the two batteries in the circuit. Your equations are fine.

You can get the third equation using Kirchoff's current law to relate i, i1, and i2. In this particular case, because of the symmetry of the circuit, you might see that i1 and i2 have to be equal.

vela

Your method is fine. You just made a mistake somewhere churning out the final answer.

Gothy

Problem 1 and 3 are done, can anyone help with the last two?

vela

There are only two loops in the circuit. Go around each one, placing and connecting the resistors as appropriate.

vela

Like I said for problem 1, you don't have to do anything differently when analyzing a circuit with two voltage sources. You seem to have a decent grasp of the material. Just get past that mental block and just do what you've been doing.

Gothy

I have tried drawing it and I only need to figure out how much Power is consumed by the B bulb. My design is an attachment below

(each resistor A,B,C, and D are 2 Ohms)
And I keep getting P_B = 18 which is not correct
Did I set up my circuit right?

Attached Thumbnails

 

Gothy

I found the Vx= 189 and I subtracted [5.8(1.15+.36)] because of voltage drops and I got 180.242.

vela

Yes, you drew the circuit correctly. Show us how you calculated P_B.

vela

This doesn't make sense. It's 189 volts relative to what?

I assume you got the 189 V by subtracting 486 V from 675 V. That 189 V would be the voltage difference between the point to the left of the 675-V battery and the point to the right of the 486-V battery.
Try applying Kirchoff's voltage law going around the loop starting from point Y, jumping straight up to point X, and then going to the left through the 675-V battery and the two resistors.

Gothy

I first calculated the equivelent resistance:
(1/(2+2) + 1/2)^-1 = 4/3
4/3 + 2 = 10/3
R_eq= 10/3

With the total resistance I found the whole circuit's current:
V=IR
I=V/R
I= 15 * 3/10= 4.5

next, I found the allocated towards the parallel circuit so I subtracted the voltage dropped from A immediately.
V_A=IR
V_A= 4.5*2 = 9

So, 6 volts allocated towards B,C, and D.
P_B=IV
P_B=V²/R_B
P_B= 36/2
P_B=18

Gothy

So what you're saying is to do 675 - 5.8*.36 - 5.8*1.15 = the potential difference? How is there a loop for the electricity to flow?
It is right but I do not understand how it works.

vela

Perfect up to here.
6 volts is the drop the series combination of B and C, not across B and C individually. Your answer would be correct for the power dissipated by D since the entire 6-volt drop appears across that resistor, but with B and C, the voltage drops across the individual resistors add up to 6 volts.

Gothy

This makes PERFECT sense! Thank you so much! I was thinking the same thing but I got confused and I forgot the square the V so instead of 4.5 I ended up with 9.

vela

Yes, that's the correct equation.

You get used to going around the actual loops in a circuit because that's how you usually get your equations, but KVL actually applies to any loop. As long as you end up in the same place as you started, the voltage differences around the loop must sum to zero.