# 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 TermUnit
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? 12 V
Battery
12 V
Battery
Resistor 1
3
Lamp
Total Circuit Resistance: 12
Circuit Current:
Solution 12 V
Battery
12 V
Battery
0 Ω
Fuse
4
Lamp
8
Solenoid
Total Circuit Resistance:
Circuit Current:

# Series Puzzle 2

What is the voltage drop across the solenoid?

How much power is the lamp using?

Solution

# Series Puzzle 3

Circuit Resistance:
Circuit Current:

## What if...?

If the switch is opened, how many watts will the motor use? 0 Ω Fuse
10
Motor
2
Resistor
0 Ω
Level Switch
12 V
Battery
Solution

# Series Puzzle 4 12 V
Battery
0 Ω
Switch
4 V
Lamp 1
6
Lamp 2
Resistor 1
10
Resistor 2
Circuit Current: 500 mA
(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?
Solution # Solutions

Want to see how we suggest solving those puzzles? Read on for a detailed walkthrough of each puzzle.

## 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

12 Ω - 3 Ω = 9 12 V
Battery
12 V
Battery
Resistor 1
3
Lamp
Total Circuit Resistance: 12
Circuit Current: ?

## 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 12 V
Battery
12 V
Battery
Resistor 1
3
Lamp
Total Circuit Resistance: 12
Circuit Current:

## 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 12 V
Battery
12 V
Battery
Resistor 1
3
Lamp
Total Circuit Resistance: 12
Circuit Current:

## 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 12 V
Battery
12 V
Battery
0 Ω
Fuse
4
Lamp
8
Solenoid
Total Circuit Resistance:
Circuit Current: ?

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

#### Step 2 12 V
Battery
12 V
Battery
0 Ω
Fuse
4
Lamp
8
Solenoid
Total Circuit Resistance:
Circuit Current:

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 12 V
Battery
12 V
Battery
0 Ω
Fuse
4
Lamp
8
Solenoid
Total Circuit Resistance:
Circuit Current:

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 12 V
Battery
12 V
Battery
0 Ω
Fuse
4
Lamp
8
Solenoid
Total Circuit Resistance:
Circuit Current:

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 0 Ω Fuse
10
Motor
2
Resistor
0 Ω
Level Switch
12 V
Battery
?
?
Circuit Resistance:Circuit Current: ?

Circuit Resistance

for total circuit resistance.

0 Ω + 10 Ω + 0 Ω + 2 Ω = 12

#### Step 2 0 Ω Fuse
10
Motor
2
Resistor
0 Ω
Level Switch
12 V
Battery
?
?
Circuit Resistance:Circuit Current:

Circuit Current

• Use Ohm's Law

Volts ÷ Ohms = Amps

12 V ÷ 24 Ω = 0.5 A

#### Step 3 0 Ω Fuse
10
Motor
2
Resistor
0 Ω
Level Switch
12 V
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

## 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 12 V
Battery
0 Ω
Switch
4 V
Lamp 1
?
6
Lamp 2
Resistor 1
10
Resistor 2
Circuit Current: 500 mA
(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

• 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

## 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 12 V
Battery
0 Ω
Switch
V
Lamp 1
6
Lamp 2
Resistor 1
10
Resistor 2

Solve for Lamp 1 Resistance

• Use Ohm's Law.

Volts ÷ Amps = Ohms

4 V ÷ 1 mA = 4