PASSPOINTPRO
PHYS 112
GENERAL PHYSICS II
KIRCHHOFF'S RULES & MULTI-LOOP CIRCUITS
100-QUESTION PRACTICE TEST
Answers, calculations & detailed explanations after every question
100 85 A-D
QUESTIONS CALCULATIONS FORMAT
Junction Rule • Loop Rule • Mesh Currents • Shared Branches • Power Checks
, FORMULA REFERENCE
Junction rule: sum I_in = sum I_out
Loop rule: sum DeltaV = 0 around any closed loop
Resistor sign: with current -> -IR; against current -> +IR
Ideal source sign: from - to + terminal -> +emf; from + to - terminal -> -emf
Two-mesh shared resistor: branch current = I1 - I2 (with consistent mesh directions)
Typical mesh equations: (R1 + Rs)I1 - Rs I2 = E1
Typical mesh equations: -Rs I1 + (R2 + Rs)I2 = E2
Ohm's law: V = IR
Resistor power: P = I^2R = VI = V^2/R
Ideal source power magnitude: |P| = |EI|
Coverage: current conservation at nodes, sign conventions, closed-loop voltage equations, two-loop mesh analysis, shared-resistor current
and voltage, resistor power, source power, and solution verification.
PassPointPro | PHYS 112 - Kirchhoff's Rules & Multi-Loop Circuits Page 2
, 1. At a junction, currents of 3 A and 1 A enter. A current of 0.7 A leaves through one branch.
What current must leave through the remaining branch?
A. 3.3 A
B. 6.6 A
C. 1.65 A
D. 4.12 A
Answer: A - 3.3 A
Explanation: Kirchhoff's junction rule is a statement of charge conservation: total current entering equals total
current leaving. Thus I = 3 + 1 - 0.7 = 3.3 A. The positive result means the assumed leaving direction is correct.
2. At a circuit node, 1.2 A and 0.9 A leave. One incoming branch carries 1 A. What current
must enter through the other incoming branch?
A. 1.1 A
B. 1.65 A
C. 2.2 A
D. 550 mA
Answer: A - 1.1 A
Explanation: Apply sum I_in = sum I_out. The unknown incoming current is I = 1.2 + 0.9 - 1 = 1.1 A. This balances
charge flow at the node.
3. A current of 1.2 A reaches a junction and splits into two branches. One branch carries 0.6
A. What current is in the second branch?
A. 1.2 A
B. 600 mA
C. 300 mA
D. 900 mA
Answer: B - 600 mA
Explanation: For a split, the incoming current equals the sum of branch currents: 1.2 = 0.6 + I. Therefore I = 1.2 -
0.6 = 0.6 A.
PassPointPro | PHYS 112 - Kirchhoff's Rules & Multi-Loop Circuits Page 3
, 4. At a junction, currents of 3 A and 0.6 A enter. A current of 1.3 A leaves through one
branch. What current must leave through the remaining branch?
A. 1.15 A
B. 2.3 A
C. 4.6 A
D. 2.88 A
Answer: B - 2.3 A
Explanation: Kirchhoff's junction rule is a statement of charge conservation: total current entering equals total
current leaving. Thus I = 3 + 0.6 - 1.3 = 2.3 A. The positive result means the assumed leaving direction is correct.
5. At a circuit node, 1.2 A and 0.6 A leave. One incoming branch carries 0.7 A. What current
must enter through the other incoming branch?
A. 1.1 A
B. 2.2 A
C. 550 mA
D. 1.65 A
Answer: A - 1.1 A
Explanation: Apply sum I_in = sum I_out. The unknown incoming current is I = 1.2 + 0.6 - 0.7 = 1.1 A. This
balances charge flow at the node.
6. A current of 2.4 A reaches a junction and splits into two branches. One branch carries 0.4
A. What current is in the second branch?
A. 4 A
B. 1 A
C. 2 A
D. 3 A
Answer: C - 2 A
Explanation: For a split, the incoming current equals the sum of branch currents: 2.4 = 0.4 + I. Therefore I = 2.4 -
0.4 = 2 A.
PassPointPro | PHYS 112 - Kirchhoff's Rules & Multi-Loop Circuits Page 4
PHYS 112
GENERAL PHYSICS II
KIRCHHOFF'S RULES & MULTI-LOOP CIRCUITS
100-QUESTION PRACTICE TEST
Answers, calculations & detailed explanations after every question
100 85 A-D
QUESTIONS CALCULATIONS FORMAT
Junction Rule • Loop Rule • Mesh Currents • Shared Branches • Power Checks
, FORMULA REFERENCE
Junction rule: sum I_in = sum I_out
Loop rule: sum DeltaV = 0 around any closed loop
Resistor sign: with current -> -IR; against current -> +IR
Ideal source sign: from - to + terminal -> +emf; from + to - terminal -> -emf
Two-mesh shared resistor: branch current = I1 - I2 (with consistent mesh directions)
Typical mesh equations: (R1 + Rs)I1 - Rs I2 = E1
Typical mesh equations: -Rs I1 + (R2 + Rs)I2 = E2
Ohm's law: V = IR
Resistor power: P = I^2R = VI = V^2/R
Ideal source power magnitude: |P| = |EI|
Coverage: current conservation at nodes, sign conventions, closed-loop voltage equations, two-loop mesh analysis, shared-resistor current
and voltage, resistor power, source power, and solution verification.
PassPointPro | PHYS 112 - Kirchhoff's Rules & Multi-Loop Circuits Page 2
, 1. At a junction, currents of 3 A and 1 A enter. A current of 0.7 A leaves through one branch.
What current must leave through the remaining branch?
A. 3.3 A
B. 6.6 A
C. 1.65 A
D. 4.12 A
Answer: A - 3.3 A
Explanation: Kirchhoff's junction rule is a statement of charge conservation: total current entering equals total
current leaving. Thus I = 3 + 1 - 0.7 = 3.3 A. The positive result means the assumed leaving direction is correct.
2. At a circuit node, 1.2 A and 0.9 A leave. One incoming branch carries 1 A. What current
must enter through the other incoming branch?
A. 1.1 A
B. 1.65 A
C. 2.2 A
D. 550 mA
Answer: A - 1.1 A
Explanation: Apply sum I_in = sum I_out. The unknown incoming current is I = 1.2 + 0.9 - 1 = 1.1 A. This balances
charge flow at the node.
3. A current of 1.2 A reaches a junction and splits into two branches. One branch carries 0.6
A. What current is in the second branch?
A. 1.2 A
B. 600 mA
C. 300 mA
D. 900 mA
Answer: B - 600 mA
Explanation: For a split, the incoming current equals the sum of branch currents: 1.2 = 0.6 + I. Therefore I = 1.2 -
0.6 = 0.6 A.
PassPointPro | PHYS 112 - Kirchhoff's Rules & Multi-Loop Circuits Page 3
, 4. At a junction, currents of 3 A and 0.6 A enter. A current of 1.3 A leaves through one
branch. What current must leave through the remaining branch?
A. 1.15 A
B. 2.3 A
C. 4.6 A
D. 2.88 A
Answer: B - 2.3 A
Explanation: Kirchhoff's junction rule is a statement of charge conservation: total current entering equals total
current leaving. Thus I = 3 + 0.6 - 1.3 = 2.3 A. The positive result means the assumed leaving direction is correct.
5. At a circuit node, 1.2 A and 0.6 A leave. One incoming branch carries 0.7 A. What current
must enter through the other incoming branch?
A. 1.1 A
B. 2.2 A
C. 550 mA
D. 1.65 A
Answer: A - 1.1 A
Explanation: Apply sum I_in = sum I_out. The unknown incoming current is I = 1.2 + 0.6 - 0.7 = 1.1 A. This
balances charge flow at the node.
6. A current of 2.4 A reaches a junction and splits into two branches. One branch carries 0.4
A. What current is in the second branch?
A. 4 A
B. 1 A
C. 2 A
D. 3 A
Answer: C - 2 A
Explanation: For a split, the incoming current equals the sum of branch currents: 2.4 = 0.4 + I. Therefore I = 2.4 -
0.4 = 2 A.
PassPointPro | PHYS 112 - Kirchhoff's Rules & Multi-Loop Circuits Page 4
