PSS 19.2: Kirchhoff’s Rules Learning Goal: To practice Problem-Solving Strategy
ID: 1329790 • Letter: P
Question
PSS 19.2: Kirchhoff’s Rules
Learning Goal:
To practice Problem-Solving Strategy 19.2 Kirchhoff’s rules.
In the circuit shown (Figure 1) , find the current in each branch.
Problem-Solving Strategy 19.2 Kirchhoff’s rules
Draw a large circuit diagram so that you have plenty of room for labels. Label all quantities, known and unknown, including an assumed direction for each unknown current and emf. If the actual direction of a particular quantity is opposite to your assumed direction, the result will come out with a negative sign.
Usually, when you label currents, it is best to use the junction rule immediately to express the currents in terms of as few quantities as possible.
SOLVE
Choose any closed loop in the network, and designate a direction (clockwise or counterclockwise) to go around the loop when applying the loop rule. The direction doesn’t have to be the same as any assumed current’s direction.
Go around the loop in the designated direction, adding potential differences as you cross them. An emf is counted as positive when you traverse it from – to + and negative when you traverse it from + to –. An IRproduct is negative if your path passes through the resistor in the same direction as the assumed current and positive if it passes through in the opposite direction. "Uphill" potential changes are always positive; "downhill" changes are always negative.
Apply Kirchhoff’s loop rule to the potential differences obtained in Step 4: V=0.
If necessary, choose another loop to get a different relation among the unknowns, and continue until you have as many independent equations as unknowns or until every circuit element has been included in at least one of the chosen loops.
Finally, solve the equations, by substitution or some other means, to determine the unknowns.
You can use this same bookkeeping system to find the potential VAB of any point A with respect to any other point B. Start at B and add the potential changes that you encounter in going from B to A, using the same sign rules as in Step 4. The algebraic sum of these changes is VAB=VAVB.
REFLECT
Always remember that when you go around a loop, adding potential differences in accordance with Kirchhoff’s loop rule, rises in potential are positive and drops in potential are negative.
SET UP
Before writing any equations, organize your information and draw appropriate diagrams.
Part A
The three currents in the circuit are defined in the figure. (Figure 2) Which of the following equations results from correct application of the junction rule for I1, I2, and I3?
Part B
Which of the following equations results from correct application of the loop rule for the circuit loop containing R1, R2, and R5? Refer to this diagram (Figure 3) to answer the question.
t
Figure 1
PSS 19.2: Kirchhoff’s Rules
Learning Goal:
To practice Problem-Solving Strategy 19.2 Kirchhoff’s rules.
In the circuit shown (Figure 1) , find the current in each branch.
Problem-Solving Strategy 19.2 Kirchhoff’s rules
Draw a large circuit diagram so that you have plenty of room for labels. Label all quantities, known and unknown, including an assumed direction for each unknown current and emf. If the actual direction of a particular quantity is opposite to your assumed direction, the result will come out with a negative sign.
Usually, when you label currents, it is best to use the junction rule immediately to express the currents in terms of as few quantities as possible.
SOLVE
Choose any closed loop in the network, and designate a direction (clockwise or counterclockwise) to go around the loop when applying the loop rule. The direction doesn’t have to be the same as any assumed current’s direction.
Go around the loop in the designated direction, adding potential differences as you cross them. An emf is counted as positive when you traverse it from – to + and negative when you traverse it from + to –. An IRproduct is negative if your path passes through the resistor in the same direction as the assumed current and positive if it passes through in the opposite direction. "Uphill" potential changes are always positive; "downhill" changes are always negative.
Apply Kirchhoff’s loop rule to the potential differences obtained in Step 4: V=0.
If necessary, choose another loop to get a different relation among the unknowns, and continue until you have as many independent equations as unknowns or until every circuit element has been included in at least one of the chosen loops.
Finally, solve the equations, by substitution or some other means, to determine the unknowns.
You can use this same bookkeeping system to find the potential VAB of any point A with respect to any other point B. Start at B and add the potential changes that you encounter in going from B to A, using the same sign rules as in Step 4. The algebraic sum of these changes is VAB=VAVB.
REFLECT
Always remember that when you go around a loop, adding potential differences in accordance with Kirchhoff’s loop rule, rises in potential are positive and drops in potential are negative.
SET UP
Before writing any equations, organize your information and draw appropriate diagrams.
Part A
The three currents in the circuit are defined in the figure. (Figure 2) Which of the following equations results from correct application of the junction rule for I1, I2, and I3?
a) I1+I2+I3=0 b) I1+I2I3=0 c) I1I2+I3=0 d) I1I2I3=0Part B
Which of the following equations results from correct application of the loop rule for the circuit loop containing R1, R2, and R5? Refer to this diagram (Figure 3) to answer the question.
Which of the following equations results from correct application of the loop rule for the circuit loop containing R1, R2, and R5?Refer to this diagram to answer the question. a) I3R5+I1R1+10V+I1R2=0 b) I3R5I1R1+10VI1R2=0 c) I3R5I1R110VI1R2=0 d) I3R5I1R1+10VI1R2=0 e) I3R5I1R110V+I1R2=0t
Figure 1
Explanation / Answer
I1 , I2 are moving towards the junction and I3 moving away from the junction
I1 + I2 = I3
I1 + I2 - I3 = 0
option (b)
++++++++++++
part(B)
starting from 10v and move in counter clock wise
+10 - I1*R1 - I3*R5 - I1*R2 = 0
-I3R5 - I1R1 + 10v - I1R2 = 0
opton ((d)
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