Comparing and solving series, parallel, and combination circuits
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Ex 6) Series Circuit
a) Find the 'equivalent resistance'
RT = R1 + R2
= 2.0 Ω + 1.0 Ω
= 3.0 ohms
b) Find the current (IT) going through this circuit
VT = ITRT
12. Volt = IT(3.0 ohms)
IT = 4.0 amps
c) Find potential drop across R1 & R2
V = IR
V1=4.0 amps(2.0 ohms) = 8.0 Volts
V2 =4.0 amps(1.0 ohms) = 4.0 Volts
Ex 7)
a) Find combined resistance (RT)
VT = IT RT
60. volts = (10 amps)RT
= 6.0 ohms
b) Find the current in R1
V1 = I1R1
60. Volts = I120. ohms
I1 = 3.0 amps
c) Find I3
V3 = VT = 60. volts
Use Ohm's Law V3 = I3R3
60. volts = I330. ohms
I3 = 2.0 amps
d) Find R2
e) Find value of the second resistor
Ex 8)
If the voltage drop across the 3 ohms resistor is 4.0 volts then the voltage drop across the 6.0 ohm resistor is
3.0 ohms : 4.0 volts
6.0 ohms : x
x = 8.0 volts
a) Find the total voltage in this series circuit
VT = V1 + V2
VT = 8.0 V + 4.0 V
= 12 V
b) Find combined resistance in this circuit
RT = 6.0 ohms + 3.0 ohms
RT = 9.0 ohms
c) Total current in this circuit? VT = ITRT
12 V = IT(9.0 ohms)
IT = 1.3 Amps
Ex 9)
a) Compare the current through each resistor
Same (Series Circuit)
IT = I1 = I2 = ...
b) Compare the voltage drops in this circuit?
20. ohm resistor has twice the potential difference (voltage)
c) Current in this circuit?
VT = ITRT 60. V = IT(30. ohms)
IT = 2.0 Amps
c) Potential difference in 20. ohm resistor?
V = IR
= 2.0 amps(20. ohms)
= 40. V
Ex 10) Find R2
V = IR
First Find RT 8 V = 1 Amp(RT)
RT = 8 ohms
Find R2
RT = R1 + R2 Since RT = 8 ohms
R = 6 ohms
Parallel Circuit Game 26
AP Circuits EMF - max. voltage of source - internal resistance causes actual V to be lower Complex Circuits
Ex) Find the potential difference between X and Y Convert parallel R to a single R 1/RT = 1/3 + 1/6 1/RT = 3/6 RT = 2 ohms
Series Circuit 4 ohms, Need V across 2 ohm R V = IR VT = ITRT 6 V = IT6 ohms IT = 1 amp
V = 2 V Ex 2)
Find equivalent R between X and Y 1/RT = 1/3 + 1/6 RT = 2 ohms
RT = 3 ohms Ex 3)
Find total current (A) 0.8 mA (B) 1.0 mA (C) 2.0 mA (D) 3.0 mA (E) 6.0 mA
Reduce parallel R to single R 1/RT=1/2000+1/6000 1/RT=3/6000+1/6000 RT = 1500 ohms
V = IR 12 V = I[4000 ohms]
(A) 0.8 mA (B) 1.0 mA (C) 2.0 mA (D) 3.0 mA (E) 6.0 mA
Enrichment
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5 quick questions — pick an answer to see if you’re right.
1. A 2.0 Ω and a 1.0 Ω resistor are in series with a 12 V battery. What is the current?
2. In that same series circuit (4.0 A through 2.0 Ω and 1.0 Ω), what is the voltage drop across the 2.0 Ω resistor?
3. In a series circuit, a 3.0 Ω resistor has a 4.0 V drop. What is the drop across a 6.0 Ω resistor in the same circuit?
4. A parallel circuit has a 60. V source and a total current of 10. A. What is the combined resistance?
5. Which statement about voltage is correct?
A 12-V battery is connected to a 4.0-Ω resistor (R1) in series with a parallel pair: 6.0 Ω (R2) and 12 Ω (R3).
What is the total current supplied by the battery?
Challenge score: 0 / 6
C1. What is the potential difference across the parallel pair?
C2. A student makes the following claim.
Place a check mark in either the Support or Refute box. Justify your response using a mathematical relationship.
C3. Substitute numerical values into a mathematical representation that can be used to determine the current through R3. Then calculate the current.
C4. Determine the total electrical energy the battery supplies to the circuit in 60. seconds.
C5. Which resistor dissipates the most power in the original circuit?
C6. Identify the evidence from this circuit that shows energy is conserved. Use the power of each resistor in your answer.