Series Circuit Puzzles Warmup
Practice, Practice, Practice!
The best way to get comfortable with electrical circuit calculations is to just do them! We've made a super easy version of the series circuit puzzles. No calculator required! You can check each answer immediately to make sure you're on the right track.
- Remember that current is the same at all points, and through each load in a series circuit.
- Assume that the components in each circuit function correctly.
- Fuses and closed switches have a resistance of 0 Ω.
In Basic Electrical Concepts, you were introduced to a couple of very handy formulas. You'll need them, as well as the concepts from the Series and Parallel Basics lesson, to solve these puzzles.
| Formula Term | Unit |
|---|---|
| Current (I) | amps (A) |
| Power (P) | watts (W) |
| Resistance (R) | ohms (Ω) |
| Voltage (V) | volts (V) |
Series Puzzle 1
Can you type in the missing values? Pay attention to the units!
What is the voltage drop across Resistor 1?
What if...?
If the value of Resistor 1 drops to 0 Ω, what would happen to the lamp?
Battery
Battery
Lamp
Battery
Battery
Fuse
Lamp
Solenoid
Series Puzzle 2
How about this circuit? Enter the missing values.
What is the voltage drop across the solenoid?
How much power is the lamp using?
Series Puzzle 3
Circuit Current:
What if...?
If the switch is opened, how many watts will the motor use?
Motor
Resistor
Level Switch
Battery
Series Puzzle 4
Battery
Switch
Lamp 1
Lamp 2
Resistor 2
(when switch is toggled as shown in schematic)
What if...?
If the switch is operated to the opposite position, what will the circuit current be?
Solutions
Want to see how we suggest solving those puzzles? Read on for a detailed walkthrough of each puzzle.
Series Puzzle 1 Walkthrough
Step 1
What is the voltage drop across Resistor 1?
Resistance of Resistor 1
- We know that the total resistance
of the circuit is 12 Ω. - Since total resistance in a series circuit
is the sum of each individual resistance,
we can subtract the other resistances
from the total to find the answer.
Total Ω - Lamp Ω = Resistor 1 Ω
12 Ω - 3 Ω = 9 Ω
Battery
Battery
Lamp
Step 2
What is the voltage drop across Resistor 1?
Circuit Current
- Use Ohm's Law
V ÷ R = I
24 V ÷ 12 Ω = 2 A
Battery
Battery
Lamp
Step 3
What is the voltage drop across Resistor 1?
Voltage Drop Across Resistor 1
- Recall that current is the same
at all points in a series circuit. - Use Ohm's Law
Resistance x Current = Voltage
9 Ω x 2 A = 18 V
Battery
Battery
Lamp
Step 4
What if...?
If the value of Resistor 1 drops to 0 Ω, what would happen to the lamp?
According to Ohm's Law,
current increases as circuit
resistance decreases.
Series Puzzle 2 Walkthrough
Step 1
Battery
Battery
Fuse
Lamp
Solenoid
What is the voltage drop across the solenoid?
How much power is the lamp using?
Total Circuit Resistance
Total series circuit resistance = sum of all circuit resistances.
Fuse Ω + Lamp Ω + Solenoid Ω = Total Ω
0 Ω + 4 Ω + 8 Ω = 12 Ω
Step 2
Battery
Battery
Fuse
Lamp
Solenoid
What is the voltage drop across the solenoid?
How much power is the lamp using?
Circuit Current
- Add the two 12 V Battery voltages together to get the total circuit voltage. (24 V)
- Add the resistances together to get the total circuit resistance. (12 Ω)
- Use Ohm's Law.
Volts ÷ Ohms = Amps
24 V ÷ 12 Ω = 2 A
Step 3
Battery
Battery
Fuse
Lamp
Solenoid
What is the voltage drop across the solenoid?
How much power is the lamp using?
Voltage Drop Across Solenoid
- Current is the same everywhere in a series circuit.
- Use Ohm's Law and the circuit current.
Ohms x Amps = Volts
8 Ω x 2 A = 16 V
Step 4
Battery
Battery
Fuse
Lamp
Solenoid
What is the voltage drop across the solenoid?
How much power is the lamp using?
How much power is the Lamp using?
- The formula for Power is found in the Basic Electrical Units lesson. (P = V x I)
- Voltage drop over the lamp can be determined by subtracting the
solenoid voltage drop from the total voltage, or using Ohm's law.
Ohms x Amps = Volts OR Vtotal - Vsolenoid = VLamp
4 Ω x 2 A = 8 V OR 24 V - 16 V = 8 V
8 V x 2 A = 16 W
Series Puzzle 3 Walkthrough
Step 1
Motor
Resistor
Level Switch
Battery
Circuit Resistance
- Add series resistances together
for total circuit resistance.
0 Ω + 10 Ω + 0 Ω + 2 Ω = 12 Ω
Step 2
Motor
Resistor
Level Switch
Battery
Circuit Current
- Use Ohm's Law
Volts ÷ Ohms = Amps
12 V ÷ 24 Ω = 0.5 A
Step 3
Motor
Resistor
Level Switch
Battery
Voltage Drops
- Use Ohm's Law
Ohms x Amps = Volts
0 Ω x 0.5 A = 0 V
20 Ω x 0.5 A= 10 V
Step 4
What if...?
If the switch is opened, how many watts will the motor use?
The circuit will become open if the switch
is opened. The motor will not operate and so
will not use power. The answer is 0 Watts.
Series Puzzle 4 Walkthrough
Step 1
Battery
Switch
Lamp 1
?
Lamp 2
Resistor 2
(when switch is toggled as shown in schematic)
Resistor 1 Resistance
- Use Ohm's Law to find the total
circuit resistance.
Volts ÷ Amps = Ohms
12 V ÷ 0.5 mA = 24 Ω
- Now subtract the other resistances
in the circuit from the total, to find
the value for Resistor 1.
24 Ω - 0 Ω - 6 Ω - 10 Ω = 8 Ω
Step 2
What if...?
If the switch is operated to the opposite position, what will the circuit current be?The circuit resistance changes when the
switch is toggled. The original current value can't be used
in this scenario.
Circuit Current
- Use Kirchoff's Voltage Law to solve
for the voltage drop across Resistor 1.
12 V - 4 V = 8 V
- Now solve for current, using Ohm's Law.
Volts ÷ Ohms = Amps
8 V ÷ 8 Ω = 1 mA
Step 3
Battery
Switch
Lamp 1
Lamp 2
Resistor 2
Solve for Lamp 1 Resistance
- Use Ohm's Law.
Volts ÷ Amps = Ohms
4 V ÷ 1 mA = 4 Ω