SOLUTION Apply the junction rule to point c . I 1 is directed into the junction,
ID: 1329653 • Letter: S
Question
SOLUTION
Apply the junction rule to point c. I1 is directed into the junction, I2 and I3 are directed out of the junction.
I1 = I2 + I3
Select the bottom loop and traverse it clockwise starting at point a, generating an equation with the loop rule.
V = Vbat + V4.0 + V9.0 = 0
6.0 V - (4.0 )I1 - (9.0 )I3 = 0
Select the top loop and traverse it clockwise from point c. Notice the gain across the 9.0 resistor because it is traversed against the direction of the current!
V = V5.0 + V9.0 = 0
-(5.0 )I2 + (9.0 )I3 = 0
Rewrite the three equations, rearranging terms and dropping units for the moment, for convenience.
(1)
I1 = I2 + I3
(2)
4.0I1 + 9.0I3 = 6.0
(3)
-5.0I2 + 9.0I3 = 0
Solve Equation (3) for I2 and substitute into Equation (1).
I2 = 1.8I3
I1 = I2 + I3 = 1.8I3 + I3 = 2.8I3
Substitute the latter expression into Equation (2) and solve for I3.
4.0 (2.8I3) + 9.0I3 = 6.0 I3= 0.30 A
Substitute I3 back into Equation (3) to get I2.
-5.0I2 + 9.0(0.30 A) = 0 I2 = 0.54 A
Substitute I3 into Equation (2) to get I1.
4.0I1 + 9.0(0.30 A) = 6.0 I1 = 0.83 A
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REMARKS Substituting these values back into the original equations verifies that they are correct, with any small discrepancies due to rounding. The problem can also be solved by first combining resistors.
QUESTION How would the answers change if the assumed directions of the currents used in solving the problem were all reversed in the figure above, but you followed the circuit paths in the same way to write down the equations based on Kirchhoff's rules? (Select all that apply.)
The sign of each current-times-resistance term would reverse.The sign used for the voltage across the battery would reverse.The sign of the resistances would reverse.The signs of the currents after solving would reverse.
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 = 4.4 .
EXERCISE HINTS: GETTING STARTED | I'M STUCK!
Use the values from PRACTICE IT to help you work this exercise. Suppose the 6.0-V battery is replaced by a battery of unknown emf, and an ammeter measures I1 = 1.4 A. Find the other two currents and the emf of the battery.
I3 = A I2 = A I1 = AExplanation / Answer
1.
Applying Kirchoff's Junction rule at point C
I1 =I2 + I3
I1-I2-I3---------------------------------------1
Applying Kirchoff's Voltage rule in bottom loop
6-4I1 -9I3=0
4I1+9I3=6-----------------------------------2
Applying Kirchoff's Voltage rule in top loop
-4.4I2+9I3=0----------------------------------3
Solving 1,2 & 3 we get
I3=0.283 A
I2=0.579 A
I3=0.862 A
-------------------------------------------------------------------------------------------------------------------------------------------------------------
2.
Given I1=1.4 A
Applying Kirchoff's Junction rule at point C
I1 =I2 + I3
I1-I2-I3---------------------------------------1
Applying Kirchoff's Voltage rule in bottom loop
E-4I1 -9I3=0
4I1+9I3=E-----------------------------------2
Applying Kirchoff's Voltage rule in top loop
-4.4I2+9I3=0
I3=(4.4/9)I2
Substituting in 1
I1-I2-(4.4/9)I2=0
1.4888I2=I1
I2=1.4/1.4888 =0.94 A
So
I3=(4.4/9)*0.94 =0.46 A
From 2
4*1.4+9*0.46=E
E=9.74 Volts
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