Kirchhoff's Rules Continued. Gustav Kirchhoff 18241887 German Physicist. Circuit Definitions. Node – any point where 2 or more circuit elements are connected together Wires usually have negligible resistance Each node has one voltage ( w.r.t . ground)
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Kirchhoff's Rules Continued
Gustav Kirchhoff
18241887
German Physicist
ProblemSolving Strategy Applying Kirchhoff’s Rules to a Circuit:
1. Assign labels and symbols to all the known and unknown quantities.
2. Assign directions to the currents in each part of the circuit. Although the assignment of current directions is arbitrary, you must stick with your original choices throughout the problem as you apply Kirchhoff ’s rules.
3. Apply the junction rule to any junction in the circuit. The rule may be applied as many times as a new current (one not used in a previously found equation) appears in the resulting equation.
4. Apply Kirchhoff’s loop rule to as many loops in the circuit as are needed to solve for the unknowns. In order to apply this rule, you must correctly identify the change in electric potential as you cross each element in traversing the closed loop. Watch out for signs!
5. Solve the equationssimultaneously for the unknown quantities, using substitution or any other method familiar to the student.
6. Check your answers by substituting them into the original equations.
Solve for the currents through each resistor
And the voltages across each resistor using
Series and parallel simplification.
The 6 and 4 ohm resistors are in series, so
are combined into 6+4 = 10Ω
The 8 and 10 ohm resistors are in parallel, so
are combined into 8∙10/(8+10) =14.4 Ω
The 10 and 4.4 ohm resistors are in series, so
are combined into 10+4 = 14.4Ω
+
I1∙14.4Ω

Writing KVL, I1∙14.4Ω – 50 v = 0
Or I1 = 50 v / 14.4Ω = 3.46 A
+34.6 v 
+
15.4 v

If I1 = 3.46 A, then I1∙10 Ω = 34.6 v
So the voltage across the 8 Ω = 15.4 v
+ 34.6 v 
+
15.4 v

If I2∙8 Ω = 15.4 v, then I2 = 15.4/8 = 1.93 A
By KCL, I1I2I3=0, so I3=I1–I2 = 1.53 A
Example (2)
Example (2):
Example (2)
Example  Kirchhoff’s Laws (4)
Example  Kirchhoff’s Laws (5)
Example (3):
Find the currents in the circuit shown in Figure by using Kirchhoff’s rules.
Solution:
Apply the junction rule to point c. I 1 is directed into
the junction, I 2 and I 3 are directed out of the junction.
Select the bottom loop, and traverse it clockwise starting
at point a, generating an equation with the loop rule:
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
Rewrite the three equations, rearranging terms and
dropping units for the moment, for convenience:
Solve eq. 3 for I2 ,
substitute into eq. 1:
Substitute I3back into eq. 3 to get I2
From eq.1
Substitute the latter expression Substitute I3 into eq. 2 to get I1
into eq. 2 and solve for I3
Homework :
Suppose the 6.0V battery is replaced by a battery of unknown emf, and an ammeter measures I1 =1.5 A. Find the other two currents and the emf of the battery.
Answers I2 =0.96 A, I3 =0.54 A, e=11 V
I1 =1.5 A
eV
Example (4):
Find I1, I2, and I3in Figure
2.5. What is the current through the 4 resistor in this circuit?
a) 0.67 A
b) 0.75 A
c) 1.0 A
d) 1.3 A
e) 1.5 A
2.6. What is the current through the 1 resistor in this circuit?
a) 2.8 A
b) 3.0 A
c) 3.4 A
d) 4.0 A
e) 4.8 A
2.7. Which one of the following equations is not correct relative to the other four equations determined by applying Kirchoff’s Rules to the circuit shown?
a) I2 = I1 + I4
b) I2 = I3 + I5
c) 6 V (8 ) I1 (5 ) I2 (4 ) I3 = 0
d) 6 V (6 ) I4 (5 ) I2 (2 ) I5 = 0
e) 6 V (8 ) I1 (6 ) I4 6 V (2 ) I5 (4 ) I3 = 0
Not a loop!
2.8.Given is the multiloop circuit on the right. Which of the following statements cannot be true:
Junction rule
Not a loop!
Upper right loop
Left loop
2.7. Consider the three resistors and the battery in the circuit shown. Which resistors, if any, are connected in parallel?
a) R1 and R2
b) R1 and R3
c) R2 and R3
d) R1 and R2and R3
e) No resistors are connected in parallel.