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Problem Solving Compound Circuits FOR SERIES CIRCUITS: V The total voltage in the circuit is the sum of the voltages across each bulb. I The current is the same anywhere in the circuit. R The total resistance in a circuit is the sum of the resistances at each bulb. FOR PARALLEL CIRCUITS: V The voltage is the same anywhere in the circuit. I The total current is the sum of the currents in each circuit. R The total resistance of a circuit follows the relationship: Data: Bulb Resistance Current Voltage A 15.0 Ω 0.33 A 5.0 V B 15.0 Ω 0.33 A 5.0 V C 15.0 Ω 0.67 A 10.0 V 1 1 1 1 = + + + ... Rt RA RB RC A C B Observations: 1. The right-hand circuit is a series circuit with bulbs A & B. 2. The left-hand circuit contains only one bulb (neither series or parallel). 3. The two circuits are in parallel. A. To determine the current near the battery (total current): First, use the series rule for current for the right-hand circuit. Current is the same in both bulbs ( 0.33 A ), and the current in the circuit is 0.33 A. Second, use the parallel rule for current for the entire circuit (right and left hand) Total current is the sum of the currents in each circuit (0.33A + 0.67 A = 1.00 A) B. To determine the voltage near the battery (total voltage): First, use the series rule for voltage for the right-hand circuit: Total voltage is the sum of voltages in the circuit. (5.0 V + 5.0 V = 10.0 V) Second, use the parallel rule for voltage for the entire circuit (right and left hand). Total voltage is the same everywhere: 10.0 V in the circuit with bulb C, and 10.0 V in the circuit with A & B. Thus the voltage at the battery is the same, 10.0 V. C. To determine the resistance near the battery (total current): First, use the series rule for the resistance for the right-hand circuit: The resistance in this circuit is the sum of the resistances for bulbs A & B (15.0 Ohms + 15.0 Ohms = 30 Ohms) Second, use the parallel rule for resistance for the entire circuit (right and left hand). 1/Rt = 1/30 + 1/15 = 1/30 + 2/30 = 3/30 = 1/10 Total resistance is 10.0 Ohms. Data: Bulb Voltage Current Resistance A 2.0 V 0.13 A 15.0 Ω B 2.0 V 0.13 A 15.0 Ω C 4.0 V 0.26 A 15.0 Ω D 6.0 V 0.40 A 15.0 Ω A B C Observations: 1. Circuit A – B is in parallel with circuit C. 2. Circuit D is in series with the circuits A-B and circuit C. D A. Current • Series rule for circuit A-B: Current is same – 0.13 A. • Parallel rule for circuit A-B with circuit C: Sum of currents (0.13 A + 0.26 A) = 0.4 A The current out of the battery splits at the junction of A-B and C. The current combines at circuit D, so that the entire current is passing through D. B. Voltage • Series rule for circuit A-B: Voltage is additive (2.0 V + 2.0 V = 4.0 V) • Since the voltage at the battery is 10.0 V, the voltage through D must add up with A-B to equal 10 V. So the voltage through D is 6.0 V. Since the voltage through D must add up with C to equal 10 V, the voltage through C must be 4.0 V. C. Resistance • Series rule for circuit A-B: Resistances are additive (15 Ohms + 15 Ohms = 30 Ohms) • Parallel rule for circuits A-B with C: 1/Rt = 1/30 + 1/15 = 3/30 = 1/10, Rt = 10 Ohms • Series rule for circuits A-B & C with D: 10 Ohms + 15 Ohms = 25 Ohms Circuit 1 Bulb Voltage Current Resistance Power A 5.0 V 0.33 A 15.0 Ω 1.65 W B 5.0 V 0.33 A 15.0 Ω 1.65 W C 5.0 V 0.33 A 15.0 Ω 1.65 W D 5.0 V 0.33 A 15.0 Ω 1.65 W Total 10.0 V 0.66 A 15.0 Ω A C B D Voltage: Series rule for A-B: 5 V + 5 V = 10 V Series rule for C-D: 5 V + 5 V = 10 V Parallel rule for total: Same as both circuits (10 V) Current: Series rule for A-B: current in circuit is equal to current in bulb – 0.33 A Series rule for C-D: same – 0.33 A Parallel rule for total: Sum of circuits (0.33 A + 0.33 A = 0.66 A) Resistance: Series rule for A-B: 15 Ohms + 15 Ohms = 30 Ohms Series rule for C-D: 30 Ohms Parallel rule for total: 1/Rt = 1/30 + 1/30 = 2/30 = 1/15 Rt = 15 Ohms Circuit 2 • • Bulb Voltage Current Resistance Power A 4.0 V 0.27 A 15.0 Ω 1.0 W B 2.0 V 0.13 A 15.0 Ω 0.26 W C 2.0 V 0.13 A 15.0 Ω 0.26 W D 4.0 V 0.27 A 15.0 Ω 1.0 W Total 10.0 V 0.27 A 37.5 Ω A B C Circuit B in parallel with C. Circuits B & C in series with A and D. Voltage in A – B – D must equal 10 V Voltage in A – C – D must equal 10 V D Current in A and D are equal. The current from A is split equally between B and C. Use parallel rule for resistance of B & C: 1/Rt = 1/15 + 1/15 = 2/15, Rt = 7.5 Ohms Use series rule for A + (B-C) + D: 15 Ohms + 7.5 Ohms + 15 Ohms = 37.5 Ohms Circuit 3 Bulb Voltage Current Resistance Power A 4.28 V 0.29 A 15.0 Ω 1.35 W B 1.43 V 0.10 A 15.0 Ω 0.15 W C 1.43 V 0.10 A 15.0 Ω 0.15 W D 1.43 V 0.10 A 15.0 Ω 0.15 W E 4.28 V 0.29 A 15.0 Ω 1.35 W Total 10.0 V 0.29 A 35.0 Ω A B Observations: • Circuits B, C, and D are parallel with each other. • This circuit is in series with A and E. Let’s look at the resistance: Use parallel rule for B-C-D: 1/Rt = 1/15 + 1/15 + 1/15 = 3/15 = Rt = 5.0 Ohms Use series rule for A + (BCD) + D: 15 Ohms + 5 Ohms + 15 Ohms = 35 Ohms C E D