EXAMPLE 18.4 Applying Kirchhoff\'s Rules GOAL Use Kirchhoff\'s rules to iind cur

Question

EXAMPLE 18.4 Applying Kirchhoff's Rules GOAL Use Kirchhoff's rules to iind currents in circuit with three currents and one allery PROBI FM Find the currents in the cirauit shown in the figur hy using Kirchhoff's SIRAIEGY There are three unknown currents In thls circult, so we must obtaln three Independent equations, which then can be solved by substitution. We can find the equations with one application of the junction rule and two applications of the oop rule. We choose junction, (Junction d gives the same equation. For the loops, we choose the bottom loop and the top loop, both shown by blue arrows, which indicate the direction we are going to traverse the circuit mathematically (not necessarlly the direction of the curent). he third loop glves an equation that can be obtalned by a linear comblnation of the other two, so it provides no additional information and isn't used SOI IITTON Apply the junction rule to point c. 1 s directed into the jurnction, 2 and 13 are directed out of the Junction. Select the hottom Inap and traverse it clockwise starting at point generating an equation with the loop ule 6.0 v-c4.0 )11" (9.0 n)/3-0 Select the top lop and traverse it clockwise from point . Notice the gain across the 9.0 resistor because it is gthe direction of the current! (1) 1 -12 (2) 4.01 9.073-6.0 (3) -5.0729.03 -1.8/3 Rewrite the three equations, rearranging terms and dropping units for the moment, for convenience. Solve Equon (3) for1 and substitute into Equation (1) Substitute the latter expression into tquation (2) and solve for3 Substiute 3 back into quation (3) to 4.0 (2.8) + 9.0, 6.0 13 0.30 A 5.0/2 + 9.0(0.30 A)-02-0-54 A 4.04 : 9.0(0.30 A)-6.011-0.83A Substitute i into Equation (2) to get LEARN MORT REMARKS Substituting these values back into the original equations verifies that they are correct, with any smiall discrepancies due to rounding, The problem can also be solved by tirst combining resistors QUESTTON How would he answers chage if the assumed ions of the currents used i solving the problem were al reversed in the figure ahove, but you followed the circuit paths in the same way to write down the equations based on Kirchhoff s rules? (Select all that apply.) Ihe sign of each current-tines, resistance term would reverse. The sign used for the voltage across the battery would reverse aThe sign of the resistances would reverse The signs of the currents aller solving would 'everse. PRACTICE IT Use the worked example above to help you solve this problem. Find the currents in the circuit as shown in the figure below by using Kirchhoff's rules. R-4.6 . EXERCISE HINIS: GETTING STARTED IM STUCK! Use the values from PRACTICE IT to help you work this exercise. Suppose the 6.0-V battery is replaced by battery of unknown emf, and n ammeter measures 11-1.2 A. Find the other two currents and the emf of the battery Need Help?

Explanation / Answer

Practice it:

Here

I1 = I2 + I3 ………………… (1)

V = Vbat + V4 + V9 = 0

6.0 – 4*I1 - 9*I3 = 0 ……………. (2)

V = V4.6 + V9 = 0

– 4.6*I2 + 9*I3 = 0………….. (3)

Solve equation (3) for I2 and put into (1),

I2 = 1.96*I3

And I1 = 1.96*I3 + I3 = 2.96*I3

Substitute later expression into equation (2) and solve for I3,

4(2.96*I3) + 9*I3 = 6 =======> I3 = 0.288 A

Put I3 back into (3) to get I2 =======> I2 = 0.563 A

Put I3 into (2) to get I1 =========> I1 = 0.851 A

Exercise:

Here equations will be

1.2 = I2 + I3 ………………… (1)

E = 4*I1 + 9*I3 ======> E = 4.8 + 9*I3 ………….(2)

– 4.6*I2 + 9*I3 = 0………….. (3)

Solve equation (3) for I2 and put into (1),

I2 = 1.96*I3

And 1.2 = 1.96*I3 + I3 = 2.96*I3

I3 = 0.405 A

Then I2 = 1.96*0.405 = 0.794 A

Then from equation (2), we gat E as

E = 4.8 + 9*0.405 = 8.445 V

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