Task Sets: Series Circuits - OpenStax Physics (High School)

A

Recall / Warm-Up

1.

Which of the following is the defining property of a series circuit?

A.

There is only one path for current to flow, so all components share the same current.

B.

Each component has the same voltage across it.

C.

The total resistance equals the smallest individual resistance.

D.

Each component can be switched on or off independently.

2.

In a series string of five holiday lights, one bulb burns out (its filament breaks, creating an open circuit). What happens to the other four bulbs?

A.

All four go out — the open circuit breaks the single current path.

B.

The other four get brighter — less resistance means more current.

C.

The other four stay on — the current finds a path around the broken bulb.

D.

Only the bulbs after the broken one go out.

3.

Kirchhoff's Voltage Law (KVL) states that in any closed loop of a circuit:

A.

The sum of all voltage drops across components equals the source voltage.

B.

The current entering a junction equals the current leaving a junction.

C.

KVL only applies to simple series circuits with one loop.

D.

The voltage drops across all resistors are equal.

B

Fluency Practice

1.

Three resistors — $R_1 = 10\ \Omega$ , $R_2 = 15\ \Omega$ , and $R_3 = 25\ \Omega$ — are connected in series. What is the equivalent resistance?

2.

Two resistors in series have a combined equivalent resistance of $R_{eq} = 45\ \Omega$ . One resistor has $R_1 = 18\ \Omega$ . What is the resistance of the other resistor?

Series circuit diagram with a 12 V battery and three resistors (10 Ω, 20 Ω, 30 Ω) connected in a single loop with current I flowing through all.

3.

A series circuit contains three resistors ( $R_1 = 10\ \Omega$ , $R_2 = 20\ \Omega$ , $R_3 = 30\ \Omega$ ) connected to a $V_{source} = 12\ \text{V}$ battery. What current flows through the circuit?

4.

Using the same circuit from fluency-3 ( $R_1 = 10\ \Omega$ , $R_2 = 20\ \Omega$ , $R_3 = 30\ \Omega$ , $V_{source} = 12\ \text{V}$ , $I = 0.20\ \text{A}$ ), what is the voltage drop across $R_3$ ?

5.

In the same series circuit ( $R_1 = 10\ \Omega$ , $R_2 = 20\ \Omega$ , $R_3 = 30\ \Omega$ , $I = 0.20\ \text{A}$ ), what is the KVL check result: do the voltage drops across all three resistors sum to $V_{source} = 12\ \text{V}$ ?

A.

Yes: $V_1 + V_2 + V_3 = 2 + 4 + 6 = 12\ \text{V}$ ✓

B.

No: the drops sum to $6\ \text{V}$ , only half the source.

C.

No: the drops cannot be calculated from the current and resistances.

D.

Yes, but only because this is a simple circuit — KVL doesn't apply in general.

C

Mixed Practice

1.

In a series circuit, $R_1 = 5\ \Omega$ is connected before $R_2 = 20\ \Omega$ (both in series with a battery). Which statement about the current through each resistor is correct?

A.

The current through $R_1$ equals the current through $R_2$ — in series, current is the same everywhere.

B.

The current through $R_1$ is larger — it's first in the circuit, so it gets more current.

C.

The current through $R_2$ is larger — bigger resistance attracts more current.

D.

The current through $R_2$ is smaller — it is "downstream" from $R_1$ which uses up some current.

2.

A series circuit has $R_1 = 12\ \Omega$ , $R_2 = 8\ \Omega$ , and $R_3 = 4\ \Omega$ connected to a $V_{source} = 6.0\ \text{V}$ battery. What is the voltage drop across $R_2$ ?

3.

In a series circuit with $R_1 = 4\ \Omega$ , $R_2 = 12\ \Omega$ , and $R_3 = 8\ \Omega$ , which resistor has the largest voltage drop?

A.

$R_2 = 12\ \Omega$ — the largest resistor has the largest voltage drop.

B.

$R_1 = 4\ \Omega$ — the first resistor in the circuit gets the most voltage.

C.

$R_2 = 12\ \Omega$ — the largest resistor gets the most current, which creates the largest drop.

D.

All three have the same voltage drop — they carry the same current.

4.

In a series circuit with $V_{source} = 15\ \text{V}$ , three resistors have voltage drops of $V_1 = 3\ \text{V}$ and $V_2 = 7\ \text{V}$ . By KVL, the voltage drop across the third resistor is ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ V. If the current in the circuit is $I = 0.50\ \text{A}$ , the resistance of $R_3$ is ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ $\Omega$ .

V_3 (volts):

R_3 (ohms):

D

Word Problems

1.

A series circuit contains three resistors: $R_1 = 10\ \Omega$ , $R_2 = 20\ \Omega$ , $R_3 = 30\ \Omega$ , connected to a $V_{source} = 12\ \text{V}$ battery.

1.

Find the equivalent resistance of the circuit.

2.

Find the current flowing through the circuit.

3.

Find the voltage drop across each resistor ( $V_1$ , $V_2$ , $V_3$ ). Enter the value for $V_2$ only.

4.

Verify with KVL: what is the sum $V_1 + V_2 + V_3$ ? (Enter the value in volts.)

2.

A voltage divider circuit has a 9.0 V battery in series with $R_1 = 10\ \Omega$ and $R_2 = ?$ . You need the voltage across $R_2$ to equal exactly 3.0 V.

What resistance must $R_2$ have? (Hint: use the voltage divider relationship $V_2 = V_{source} \times R_2 / R_{eq}$ .)

E

Error Analysis

1.

Student's explanation:

"In a series circuit with $R_1 = 10\ \Omega$ and $R_2 = 20\ \Omega$ connected to a 6 V battery, the current entering $R_1$ is $I = 6/10 = 0.6\ \text{A}$ . After passing through $R_1$ , some current is converted to heat, so less current enters $R_2$ . The current through $R_2$ is $I = (6-6)/20 = 0/20 = 0\ \text{A}$ ."

A student explains current flow in a series circuit. Identify the error.

A.

The student used the wrong formula for current — it should be $I = V \times R$ .

B.

The student confused energy and charge. Resistors convert energy to heat, but do not consume charge. The same current flows through all series components. The correct current is $I = 6/(10+20) = 0.20\ \text{A}$ through both resistors.

C.

The student is right — after a resistor converts energy to heat, there is no current left for the next component.

D.

The calculation for $R_1$ is correct, but $R_2$ should use a different formula.

2.

Student's reasoning:

"Since $R_2 = 25\ \Omega$ is bigger than $R_1 = 5\ \Omega$ , the larger resistor $R_2$ must draw more current. So the current through $R_2$ is five times the current through $R_1$ ."

A student analyzes a series circuit with $R_1 = 5\ \Omega$ and $R_2 = 25\ \Omega$ connected to a 15 V battery. Read their reasoning and find the error.

A.

The student is right — bigger resistors attract more current.

B.

The student is wrong. In a series circuit, all components carry the same current. The larger resistor gets more voltage, not more current: $V_2 = I \times R_2$ is larger because $R_2$ is larger, but $I$ is the same for both.

C.

The student is right that $R_2$ draws more current, but the ratio should be $R_1/R_2$ , not $R_2/R_1$ .

D.

Current is not defined for individual resistors — only for the whole circuit.

F

Challenge

1.

A series circuit has a 24 V battery, $R_1 = 6\ \Omega$ , and an unknown $R_2$ . The voltage across $R_1$ is measured to be $V_1 = 9.0\ \text{V}$ . Find the resistance of $R_2$ .

2.