Kirchhoff's Voltage Law

jena

Hi,

My question:

The batteries have emfs of E1=9.0 V and E2= 12.0 V and the resistors have values of R1=25 ohms, R2=18 ohms, and R3=35 ohms.Assuming each battery has internal resistance r=1.0 ohms find the direction and magnitude of the currents.


____R1______
|.....................|
E1...................|
|_____R2______|
|.....................|
E2...................R3
|_____________|


For this problem could I use this equation to find the currents.

V(ab)=E-Ir and I= V/R

Is this correct

Thank You

Tom Mattson

Since you have not labeled nodes "a" and "b", it's not possible to say.

jena

Please disregard the ab nodes.

Thank You

Tom Mattson

OK, even if we disregard the "ab", you still aren't going to solve this with a canned equation. You have to apply Kirchhoff's laws to the problem. I think that the easiest way to solve the problem is by superposition. That is, solve the problem once by deactivating E1 and solve it again by deactivating E2 (deactivated voltage sources are just short circuits). Then you can find the currents in each case and add the results together.

jena

So it would be

I=V/R

I(1)=E1/(R1+R2)

I(2)=E2/(R1+R3)

to help find the currents needed

Tom Mattson

No, you have to apply Kirchhoff's laws. This requires some analysis beyond simple plug-and-chug.

As I was explaining, I think the easiest way to do this problem would be to look at the following two circuits:

____R1______
|.....................|
E1...................|
|_____R2______|
|.....................|
|....................R3
|_____________|

and

____R1______
|.....................|
|.....................|
|_____R2______|
|.....................|
E2...................R3
|_____________|

Find the current through each resistor for both circuits. Then you can add up the results for each circuit to get the current through each resistor when both sources are activated. This is called the principle of superposition, and it works because the element law for resistors (V=IR) is linear.

jena

Hi,

I have the currents:

I(1)=.68 A right
I(2)=.31 A left
I93)= .18 A up

what's next

Thank You

lightgrav

Jena,

if 0.68 C/sec flows into the solder-spot on the right
(where R1 joins R2 and joins R3),
then 0.68 C/sec must flow OUT of that node.
(you have (0.31 - 0.18)[Amp] = 0.13 C/sec out.)

Why would I3 flow up? only if E2 is + at bottom.
But then I2 must be large.

You Did simplify with series and parallel R's first?

jena

Hi,

I tried it again and this time I set equations to find the currents using the loop rule.

Loop 1:

E1- I1(R1+r)-I2(R2), where is the r, is the internal resistance

____R1___>___
|....................|
r.....................|
E1..................V
|_____R2_<___|


Loop 2:
-E2+I2(R2)+I3(R3+r), again where r is the internal resistance

____R2___<___
^...................|
|....................|
E2..................^
|_r____R2_>___|

Is this correct

Thank You

jena

Hi,

I think I got it now, the work is below.

1. Set up loops

Loop 1: E1-(R1+r)(I1)-R2(I2)=0
9V-(25ohms+1ohms)(I1)-(18ohms)(I2)=0
9V-(26ohms)(I1)-(18ohms)(I2)=0

Loop 2: E2-(R3+r)(I3)-R2(I2)=0
12V-(35ohms+1ohms)(I1)-(18ohms)(I2)=0
12V-(36ohms)(I3)-(18ohms)(I2)=0

2. Substitution: Add Junctin equation in: I2= I1+I3

Loop 1: 9V-(26ohms)(I1)-(18ohms)(I2)=0
9V-(26ohms)(I1)-(18ohms)(I1+I3 )=0
9V-(44ohms)(I1)-(18ohms)(I3 )=0
-(44ohms)(I1)-(18ohms)(I3 )=-9V

Loop 2: 12V-(36ohms)(I3)-(18ohms)(I2)=0
12V-(36ohms)(I3)-(18ohms)(I1+I3 )=0
12V-(54ohms)(I3)-(18ohms)(I1 )=0
-(54hms)(I3)-(18ohms)(I1)=-12V

3. Add/Subtract Loop 1 and Loop 2. Solve for I3

Loop 1:18( -(46ohms)(I1)-(18ohms)(I3 )=-9V)
Loop 2: -44( -(54hms)(I3)-(18ohms)(I1)=-12V)

I3=(366 V)/(2052ohms)
I3=.178 A

4. Solve for I2, using loop 2

Loop 2:12V-(36ohms)(I3)-(18ohms)(I2)=0
12V-(36ohms)(.178 A)-(18ohms)(I2)=0
I2=.309 A

5. Sove for I1, using junction equation
Junction equation: I2=I1+I3
.309A= I1 +.178A
.309A -.178 A= I1
I1 =.131 A

Is this the correct way to do it

Thank You