## Kirchhoff's Voltage Law

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
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 Quote by jena For this problem could I use this equation to find the currents. V(ab)=E-Ir and I= V/R Is this correct
Since you have not labeled nodes "a" and "b", it's not possible to say.
 Please disregard the ab nodes. Thank You

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## Kirchhoff's Voltage Law

 Quote by jena For this problem could I use this equation to find the currents. V(ab)=E-Ir and I= V/R
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.
 So it would be I=V/R I(1)=E1/(R1+R2) I(2)=E2/(R1+R3) to help find the currents needed
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus 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.
 Hi, I have the currents: I(1)=.68 A right I(2)=.31 A left I93)= .18 A up what's next Thank You
 Recognitions: Homework Help 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?
 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
 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