🎯⭐ INTERACTIVE LESSON

Series and Parallel Circuits

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Series and Parallel Circuits - Complete Interactive Lesson

Part 1: Series Circuits

🔗 Series Circuits

Part 1 of 7 — One Path, Shared Current

When resistors are connected end-to-end with only one path for current to flow, they form a series circuit. Understanding series circuits is the foundation for all circuit analysis.

In this part you'll learn:

How to identify series connections
Why current is the same through every series element
How to calculate total resistance in series
How voltage divides across series resistors
What happens when one element breaks

What Makes a Series Circuit?

Resistors are in series when they are connected end-to-end so that there is only one path for current to flow.

Key Properties of Series Circuits

Property	Rule
Current	Same through every element: $I_1 = I_2 = I_3 = \ldots = I$

Why Is the Current the Same?

Think of water flowing through a single pipe with narrow sections. The same volume of water per second must pass through every point — there's nowhere else for it to go!

Similarly, in a series circuit, charge cannot accumulate at any junction between resistors. The same current $I$ flows through every component.

Total Resistance

Since each resistor opposes current flow, adding resistors in series makes it harder for current to flow:

$R_{\text{total}} = R_1 + R_2 + R_3 + \cdots$

The total resistance is always greater than the largest individual resistance.

Series Circuit Concepts

Voltage Division in Series

Since the current is the same through every resistor, the voltage across each one is given by Ohm's law:

$V_k = I \cdot R_k$

A larger resistor gets a larger share of the total voltage.

Voltage Divider Rule

The voltage across resistor in a series circuit is:

Series Circuit Calculation Drill

A 24 V battery is connected to three resistors in series: $R_1 = 4 \; \Omega$ , $R_2 = 8 \; \Omega$ , .

Advanced Series Drill

Two resistors are connected in series to a 20 V battery. The voltage across the first resistor is 8 V, and the current through the circuit is 0.5 A.

Voltage across the second resistor (in V)
Resistance of the first resistor (in $\Omega$ )
Resistance of the second resistor (in $\Omega$ )

Exit Quiz — Series Circuits

Part 2: Parallel Circuits

🔀 Parallel Circuits

Part 2 of 7 — Multiple Paths, Shared Voltage

When resistors are connected so that both ends of each resistor share the same two nodes, current has multiple paths to follow. This is a parallel circuit — and the rules are very different from series.

In this part you'll learn:

How to identify parallel connections
Why voltage is the same across every parallel element
How to calculate total resistance in parallel
How current divides among parallel branches
Why parallel resistance is always less than the smallest individual resistor

What Makes a Parallel Circuit?

Resistors are in parallel when they are connected between the same two nodes — each resistor provides a separate path for current.

Key Properties of Parallel Circuits

Property	Rule
Voltage	Same across every branch: $V_1 = V_2 = V_3 = \ldots = V$

Part 3: Combination Circuits

🧩 Combination Circuits

Part 3 of 7 — Series and Parallel Together

Most real circuits aren't purely series or purely parallel — they contain combinations of both. The strategy is to identify series and parallel groups, simplify step by step, then work backward to find individual voltages and currents.

In this part you'll learn:

How to identify series and parallel groups in a complex circuit
The step-by-step reduction method
How to work backward to find voltages and currents for each resistor
Multi-step circuit analysis problems

Circuit Reduction Strategy

Step-by-Step Method

Identify groups of resistors that are purely in series or purely in parallel
Replace each group with a single equivalent resistor
Repeat until you have a single equivalent resistance
Find total current using $I = V/R_{\text{total}}$

Part 4: Voltage & Current Dividers

⚡ Voltage Dividers & Current Dividers

Part 4 of 7 — Practical Circuit Design Tools

Voltage dividers and current dividers are essential building blocks in electronics. They let you create specific voltages or steer currents without complex components — just resistors!

In this part you'll learn:

The voltage divider formula and when to use it
The current divider formula and when to use it
How potentiometers work as adjustable voltage dividers
Practical applications in circuit design

The Voltage Divider

A voltage divider is two resistors in series that produce an output voltage that is a fraction of the input voltage.

The Formula

For two resistors $R_1$ and $R_2$ in series, connected to source voltage :

Part 5: Power in Circuits

💡 Power in Circuits

Part 5 of 7 — Energy Dissipation in Series & Parallel

Every resistor converts electrical energy into heat. Understanding how power distributes among resistors is crucial for circuit design (and for the AP exam!).

In this part you'll learn:

Three forms of the power equation
Power distribution in series circuits
Power distribution in parallel circuits
Conservation of energy: total power equals source power

Power Equations

The power dissipated by a resistor can be calculated three ways:

$P = IV \qquad P = I^2 R \qquad P = \frac{V^2}{R}$

Part 6: Capacitors in Circuits

🔋 Capacitors in Series & Parallel

Part 6 of 7 — The Rules Are Flipped!

Capacitors combine in series and parallel using rules that are the opposite of resistors. This is one of the most common sources of mistakes on the AP exam — so pay close attention!

In this part you'll learn:

How capacitors combine in parallel (add directly)
How capacitors combine in series (reciprocals add)
Why the rules are "opposite" to resistors
Energy stored in capacitors: $U = \frac{1}{2}CV^2$

Part 7: Synthesis & AP Review

🎯 Synthesis & AP Review

Part 7 of 7 — Putting It All Together

This final part integrates everything from Parts 1–6. You'll develop a systematic circuit analysis strategy, review the most common AP mistakes, and tackle AP-style problems.

In this part you'll learn:

A step-by-step strategy for any circuit problem
Common mistakes that cost points on the AP exam
How to approach AP free-response circuit questions
A comprehensive mastery quiz

Circuit Analysis Strategy

Step-by-Step Approach

Draw and label — Redraw the circuit neatly. Label all resistors/capacitors and the source.
Identify topology — Find series and parallel groups.
Simplify — Reduce combination groups one step at a time.
Solve for totals — Find $R_{\text{total}}$ (or $_{}$ ), then or .

$V_k = V_{\text{source}} \cdot \frac{R_k}{R_{\text{total}}}$

Three resistors in series: $R_1 = 10 \; \Omega$ , $R_2 = 20 \; \Omega$ , $R_3 = 30 \; \Omega$ connected to $V = 12$ V.

$R_{\text{total}} = 10 + 20 + 30 = 60 \; \Omega$

$I = \frac{V}{R_{\text{total}}} = \frac{12}{60} = 0.2 \text{ A}$

Resistor	Voltage
$R_1 = 10 \; \Omega$	$V_1 = 0.2 \times 10 = 2$ V
$R_2 = 20 \; \Omega$	$V_2 = 0.2 \times 20 = 4$ V
$R_3 = 30 \; \Omega$	$V_3 = 0.2 \times 30 = 6$ V

Check: $2 + 4 + 6 = 12$ V ✓ — voltages add up to the source voltage.

If any resistor in a series circuit burns out (open circuit), current drops to zero — the entire circuit stops. This is like old-fashioned Christmas lights: one bulb out, all out!

R_3 = 12 \; \Omega

Total resistance $R_{\text{total}}$ (in $\Omega$ )
Current through the circuit (in A)
Voltage across $R_2$ (in V)
Voltage across $R_3$ (in V)

Why Is the Voltage the Same?

Each parallel resistor is directly connected to the same two nodes (same two wires). By definition, the potential difference between those two nodes is the same no matter which path you take — it's the same voltage.

Each additional parallel path gives current another way to flow, so the total resistance decreases:

$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots$

For two resistors in parallel, there's a useful shortcut:

$R_{\text{total}} = \frac{R_1 \cdot R_2}{R_1 + R_2}$

Key fact: $R_{\text{total}}$ is always less than the smallest individual resistance.

Parallel Circuit Concepts

Current Division in Parallel

Since each branch sees the same voltage, the current through each branch depends on its resistance:

$I_k = \frac{V}{R_k}$

A smaller resistance carries a larger current — current prefers the easy path!

Example

Two resistors in parallel: $R_1 = 4 \; \Omega$ and $R_2 = 12 \; \Omega$ , connected to a V battery.

$I_1 = \frac{12}{4} = 3 \text{ A} \qquad I_2 = \frac{12}{12} = 1 \text{ A}$

$I_{\text{total}} = 3 + 1 = 4 \text{ A}$

Check: $R_{\text{total}} = \frac{4 \times 12}{4 + 12} = \frac{48}{16} = 3 \; \Omega$ . A ✓

If One Branch Breaks...

Unlike series circuits, if one branch in a parallel circuit opens, the other branches continue to operate. Current simply flows through the remaining paths. This is why household circuits are wired in parallel — one light burning out doesn't affect the others!

Parallel Circuit Calculation Drill

A 30 V battery is connected to three resistors in parallel: $R_1 = 10 \; \Omega$ , $R_2 = 15 \; \Omega$ , $R_3 = 30 \; \Omega$ .

Current through $R_1$ (in A)
Current through $R_2$ (in A)
Current through $R_3$ (in A)

Special Case: Identical Resistors in Parallel

When $n$ identical resistors of resistance $R$ are connected in parallel:

$R_{\text{total}} = \frac{R}{n}$

This is a very useful shortcut!

$n$ identical resistors	$R_{\text{total}}$
2 in parallel	$R/2$
3 in parallel	$R /$

Example

Four $100 \; \Omega$ resistors in parallel:

$R_{\text{total}} = \frac{100}{4} = 25 \; \Omega$

Each resistor carries one-quarter of the total current.

Exit Quiz — Parallel Circuits

Work backward — expand each group and use series/parallel rules to find individual

V

and

I

values

How to Tell Series from Parallel

Series: Two resistors are in series if all the current through one must also pass through the other (no branching between them)
Parallel: Two resistors are in parallel if they share the same two nodes (same start point and same end point)

Example: Three-Resistor Combination

Consider: $R_1 = 6 \; \Omega$ in series with the parallel combination of $R_2 = 4 \; \Omega$ and $R_3 = 12 \; \Omega$ . Battery: $V = 24$ V.

Step 1: Find the parallel combination: $R_{23} = \frac{R_2 \cdot R_3}{R_2 + R_3} = \frac{4 \times 12}{4 + 12} = \frac{48}{16} = 3 \; \Omega$

Step 2: Now $R_1$ and $R_{23}$ are in series: $R_{\text{total}} = R_1 + R_{23} = 6 + 3 = 9 \; \Omega$

Step 3: Total current: $I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{24}{9} = \frac{8}{3} \approx 2.67 \text{ A}$

Step 4: Work backward:

$V_1 = I \cdot R_1 = \frac{8}{3} \times 6 = 16$ V
$V_{23} = I \cdot R_{23} = \frac{8}{3} \times 3 = 8$ V (same voltage across and )
$I_2 = V_{23}/R_2 = 8/4 = 2$ A
$I_3 = V_{23}/R_3 = 8/12 = 2/3$ A

Check: $I_2 + I_3 = 2 + 2/3 = 8/3$ A ✓ and $V_1 + V_{23} = 16 + 8 = 24$ V ✓

Identifying Series & Parallel

Combination Circuit Drill

$R_1 = 5 \; \Omega$ is in series with the parallel combination of $R_2 = 10 \; \Omega$ and $R_3 = 40 \; \Omega$ . The battery provides $V = 20$ V.

Equivalent resistance of $R_2$ and $R_3$ in parallel (in $\Omega$ )
Total circuit resistance (in $\Omega$ )

Round all answers to 3 significant figures.

Four-Resistor Combination

Consider a more complex circuit:

$R_1 = 2 \; \Omega$ and $R_2 = 6 \; \Omega$ are in parallel. This parallel combination is in series with $R_3 = 4 \; \Omega$ and $R_4 = 3 \; \Omega$ . Battery: $V = 18$ V.

Step 1: Simplify the parallel pair

$R_{12} = \frac{2 \times 6}{2 + 6} = \frac{12}{8} = 1.5 \; \Omega$

Step 2: Add all series resistances

$R_{\text{total}} = R_{12} + R_3 + R_4 = 1.5 + 4 + 3 = 8.5 \; \Omega$

Step 3: Total current

$I_{\text{total}} = \frac{18}{8.5} \approx 2.12 \text{ A}$

Step 4: Voltages and currents

$V_{12} = 2.12 \times 1.5 \approx 3.18$ V
$V_3 = 2.12 \times 4 \approx 8.47$ V

Check: $3.18 + 8.47 + 6.35 = 18$ V ✓ and $1.59 + 0.53 \approx 2.12$ A ✓

Exit Quiz — Combination Circuits

$V_{\text{out}} = V_{\text{in}} \cdot \frac{R_2}{R_1 + R_2}$

where $V_{\text{out}}$ is measured across $R_2$ (the "bottom" resistor).

From the series voltage division rule: $V_2 = I \cdot R_2 = \frac{V_{\text{in}}}{R_1 + R_2} \cdot R_2 = V_{\text{in}} \cdot \frac{R_2}{R_1 + R_2}$

$V_{\text{in}} = 9$ V, $R_1 = 3 \; \text{k}\Omega$ , $R_2 = 6 \; \text{k}\Omega$ :

$V_{\text{out}} = 9 \times \frac{6}{3 + 6} = 9 \times \frac{6}{9} = 6 \text{ V}$

If $R_2 \gg R_1$ : $V_{\text{out}} \approx V_{\text{in}}$ (most voltage across $R_2$ )
If $R_2 \ll R_1$ : $V_{\text{out}} \approx 0$ (most voltage across )
If $R_1 = R_2$ : $V_{\text{out}} = V_{\text{in}}/2$ (voltage splits equally)

Voltage Divider Drill

A voltage divider with $R_1 = 2 \; \text{k}\Omega$ and $R_2 = 8 \; \text{k}\Omega$ is connected to 10 V. What is $V_{\text{out}}$ across $R_2$ ? (in V)
You need $V_{\text{out}} = 3$ V from a 12 V source using a voltage divider. If $R_2 = 1 \; \text{k}\Omega$ , what must be? (in k)
A voltage divider uses $R_1 = R_2 = 5 \; \text{k}\Omega$ with a 20 V input. What is $V_{\text{out}}$ ? (in V)

Potentiometers

A potentiometer (or "pot") is an adjustable voltage divider. It's a single resistor with a sliding contact (wiper) that divides it into two parts.

How It Works

Total resistance: $R_{\text{total}}$

The wiper position determines the split:

Upper portion: $R_1 = (1-x) \cdot R_{\text{total}}$
Lower portion: $R_2 = x \cdot R_{\text{total}}$

where $x$ is the fractional position (0 to 1) of the wiper from bottom to top.

$V_{\text{out}} = V_{\text{in}} \cdot \frac{x \cdot R_{\text{total}}}{R_{\text{total}}} = x \cdot V_{\text{in}}$

So a potentiometer gives you a continuously adjustable output from 0 to $V_{\text{in}}$ .

Applications

Volume knobs on audio equipment
Dimmer switches for lights
Joysticks in game controllers
Sensor readout circuits (thermistors, photoresistors)

The Current Divider

A current divider is two resistors in parallel that split the incoming current.

The Formula

For two parallel resistors carrying total current $I_{\text{total}}$ :

$I_1 = I_{\text{total}} \cdot \frac{R_2}{R_1 + R_2}$

$I_2 = I_{\text{total}} \cdot \frac{R_1}{R_1 + R_2}$

Key Insight — "Opposite" from Voltage Divider!

Notice the "flip": in the current divider, $I_1$ depends on $R_2$ (not $R_1$ ). The smaller resistor gets the share of the current.

Example

$I_{\text{total}} = 6$ A splits between $R_1 = 4 \; \Omega$ and in parallel:

$I_1 = 6 \times \frac{12}{4 + 12} = 6 \times \frac{12}{16} = 4.5 \text{ A}$

$I_2 = 6 \times \frac{4}{4 + 12} = 6 \times \frac{4}{16} = 1.5 \text{ A}$

The $4 \; \Omega$ resistor (smaller) gets the larger current (4.5 A). ✓

Current Divider Drill

A total current of 10 A enters a node and splits between $R_1 = 6 \; \Omega$ and $R_2 = 4 \; \Omega$ in parallel.

Current through $R_1$ (in A)
Current through $R_2$ (in A)
Voltage across the parallel combination (in V)

Exit Quiz — Dividers

All three are equivalent (just substitute $V = IR$ or $I = V/R$ ).

Known Quantities	Best Formula
$I$ and $V$	$P = IV$
$I$ and $R$	$P = I^2 R$
$V$ and $R$	$P = V^2/R$

$[P] = \text{Watts (W)} = \frac{\text{J}}{\text{s}} = \text{A} \cdot \text{V}$

Conservation of Energy

The total power delivered by the battery equals the total power dissipated by all resistors:

$P_{\text{source}} = P_1 + P_2 + P_3 + \cdots$

This is just conservation of energy — the battery's energy output per second equals the total heat output per second.

Power Concept Check

Power in Series Circuits

In series, the current $I$ is the same through every resistor:

$P_k = I^2 R_k$

The larger resistor dissipates more power in a series circuit (since $I$ is constant and $P \propto R$ ).

Example: Series

$R_1 = 2 \; \Omega$ and $R_2 = 6 \; \Omega$ in series, V:

$R_{\text{total}} = 8 \; \Omega, \qquad I = 16/8 = 2 \text{ A}$

$P_1 = (2)^2 \times 2 = 8 \text{ W}$ $P_{2} = (2$

Check: $P_{\text{source}} = IV = 2 \times 16 = 32$ W ✓

Power in Parallel Circuits

In parallel, the voltage $V$ is the same across every resistor:

$P_k = \frac{V^2}{R_k}$

The smaller resistor dissipates more power in a parallel circuit (since $V$ is constant and $P \propto 1/R$ ).

Example: Parallel

$R_1 = 3 \; \Omega$ and $R_2 = 6 \; \Omega$ in parallel, V:

$P_1 = \frac{12^2}{3} = 48 \text{ W} \qquad P_2 = \frac{12^2}{6} = 24 \text{ W}$

$P_{\text{total}} = 48 + 24 = 72 \text{ W}$

Check: $R_{\text{total}} = 3 \times 6/(3+6) = 2 \; \Omega$ . A. W ✓

Series vs. Parallel Summary

	Series	Parallel
Same quantity	Current $I$	Voltage $V$
Best power formula	$P = I^2 R$

Power Distribution Drill

A 24 V battery is connected to $R_1 = 4 \; \Omega$ in series with the parallel combination of $R_2 = 6 \; \Omega$ and $R_3 = 12 \; \Omega$ .

Total resistance of the circuit (in $\Omega$ )
Total current (in A)
Power dissipated by $R_1$ (in W)
Power dissipated by $R_2$ (in W)
Total power delivered by the battery (in W)

Lightbulb Brightness & Power

On the AP exam, questions often ask about brightness of identical lightbulbs in different configurations. Brightness is proportional to power dissipated.

Identical Bulbs (each resistance $R$ )

Two bulbs in series (battery voltage $V$ ): $I = \frac{V}{2R}, \qquad P_{\text{each}} = I^2 R = \frac{V^2}{4R}$

Two bulbs in parallel (battery voltage $V$ ): $P_{\text{each}} = \frac{V^2}{R}$

Ratio: Each parallel bulb is 4× brighter than each series bulb!

Why?

In parallel, each bulb gets the full battery voltage. In series, each bulb gets only half.

$\frac{P_{\text{parallel}}}{P_{\text{series}}} = \frac{V^2/R}{V^2/(4R)} = 4$

Exit Quiz — Power in Circuits

Charge distribution in series and parallel

Capacitors in Parallel

When capacitors are in parallel, they share the same voltage. Each capacitor stores charge independently:

$Q_1 = C_1 V, \quad Q_2 = C_2 V, \quad Q_3 = C_3 V$

Total charge: $Q_{\text{total}} = Q_1 + Q_2 + Q_3 = (C_1 + C_2 + C_3)V$

Parallel Capacitance Formula

$C_{\text{parallel}} = C_1 + C_2 + C_3 + \cdots$

Capacitors in parallel ADD directly — just like resistors in series!

Why?

Connecting capacitors in parallel effectively increases the total plate area. More area → more charge storage → more capacitance.

Example

$C_1 = 2 \; \mu\text{F}$ , $C_2 = 3 \; \mu\text{F}$ , in parallel:

$C_{\text{total}} = 2 + 3 + 5 = 10 \; \mu\text{F}$

Capacitors in Series

When capacitors are in series, they all store the same charge $Q$ (just like current is the same in series resistors). The voltage divides:

$V_{\text{total}} = V_1 + V_2 + V_3 = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3}$

Series Capacitance Formula

$\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \cdots$

Capacitors in series use the RECIPROCAL rule — just like resistors in parallel!

For two capacitors in series:

$C_{\text{series}} = \frac{C_1 \cdot C_2}{C_1 + C_2}$

Key Fact

$C_{\text{series}}$ is always less than the smallest individual capacitance.

Why?

Connecting capacitors in series effectively increases the plate separation. More separation → less capacitance.

Example

$C_1 = 6 \; \mu\text{F}$ and $C_2 = 3 \; \mu\text{F}$ in series:

$C_{\text{series}} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2 \; \mu\text{F}$

Resistors vs. Capacitors: The Flip

	Resistors	Capacitors
Series	$R = R_1 + R_2$ (add)	$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$ (reciprocal)
Parallel	$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ (reciprocal)
Series: same...	Current	Charge
Parallel: same...	Voltage	Voltage

Memory Trick

Resistors: Reciprocal in paRallel Capacitors: the rules are Completely flipped!

Energy Stored in a Capacitor

$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$

All three forms are equivalent, related by $Q = CV$ .

Capacitor Combination Concepts

Capacitor Combination Drill

$C_1 = 4 \; \mu\text{F}$ and $C_2 = 12 \; \mu\text{F}$ are connected in series across a 16 V battery.

Total capacitance (in $\mu$ F)
Charge stored on each capacitor (in $\mu$ C)
Voltage across $C_1$ (in V)
Voltage across $C_2$ (in V)

Exit Quiz — Capacitors

Expand backward — Work back through each simplification step, applying:

Series: same $I$ , voltages add
Parallel: same $V$ , currents add

Verify — Check that voltages around any loop sum to zero, currents at any node balance, and total power balances.

Quantity	Series	Parallel
$R$	$R_1 + R_2$	$\frac{R_1 R_2}{R_1 + R_2}$
$C$	$\frac{C_1 C_2}{C_1 + C_2}$
Same...	$I$ (resistors), $Q$ (capacitors)	$V$ (both)
Splits...	$V$	$I$ (resistors), $Q$ (capacitors)

Common AP Mistakes

❌ Mistake 1: Confusing Resistor and Capacitor Rules

	Series adds	Reciprocal adds
Resistors	✅ Series	✅ Parallel
Capacitors	✅ Parallel	✅ Series

They're opposite! The #1 error is applying resistor series rules to capacitors in series.

❌ Mistake 2: "Bigger R = More Power" Always

In series: yes, bigger $R$ → more power ( $P = I^2 R$ )
In parallel: no, bigger $R$ → less power ( $P = V^2/R$ )

The relationship depends on whether current or voltage is shared!

❌ Mistake 3: Forgetting to Work Backward

Finding $R_{\text{total}}$ is only half the problem. You must expand back to find individual voltages and currents.

❌ Mistake 4: Assuming Equal Distribution

In series, voltage does NOT split equally (unless $R$ values are equal). In parallel, current does NOT split equally (unless $R$ values are equal).

❌ Mistake 5: Adding Reciprocals Incorrectly

$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} \neq \frac{1}{R_1 + R_2}$

Don't forget to take the final reciprocal!

AP Free-Response Strategy

AP Physics 2 circuit FRQs typically ask you to:

Part (a): Rank or Compare

"Rank the brightness of the bulbs" or "Compare the current through each resistor."

Strategy: Identify the topology, determine which quantities are shared, then compare.

Part (b): Calculate

"Determine the current through $R_2$ " or "Calculate the total power."

Strategy: Reduce the circuit, find totals, expand backward.

Part (c): Predict a Change

"A switch opens, disconnecting $R_3$ . What happens to the brightness of $R_1$ ?"

Strategy:

Analyze the circuit before the change
Analyze the circuit after the change
Compare the quantity of interest

Part (d): Justify with Physics

"Explain your reasoning using principles of circuit analysis."

Strategy: State the relevant rule (Kirchhoff's laws, Ohm's law, series/parallel properties) and connect it logically to your answer.

Synthesis Problem

A circuit has a 36 V battery connected to $R_1 = 12 \; \Omega$ in series with the parallel combination of $R_2 = 8 \; \Omega$ and $R_3 = 24 \; \Omega$ .

$R_{\text{total}}$ of the circuit (in $\Omega$ )
Total current from the battery (in A)
Power dissipated by $R_1$ (in W)
Current through (in A)

Round all answers to 3 significant figures.

Mastery Quiz — Series & Parallel Circuits

Total current from the battery (in A)

Total resistance of the circuit (in

\Omega

)

Total current from the battery (in A)

Voltage across

R_1

(in V)

Voltage across the parallel group (in V)

Current through

R_2

(in A)

V_4 = 2.12 \times 3 \approx 6.35

I_1 = 3.18/2 \approx 1.59

I_2 = 3.18/6 \approx 0.53

R_2 = 12 \; \Omega

P_{2} = (2)^{2} \times 6 = 24 W

P_{\text{total}} = 8 + 24 = 32 \text{ W}

P = 6 \times 12 = 72

C_3 = 5 \; \mu\text{F}

Total energy stored (in

\mu

Total power delivered by the battery (in W)

Series and Parallel Circuits

Series and Parallel Circuits - Complete Interactive Lesson

Part 1: Series Circuits

🔗 Series Circuits

What Makes a Series Circuit?

Key Properties of Series Circuits

Why Is the Current the Same?

Total Resistance

Voltage Division in Series

Voltage Divider Rule

Part 2: Parallel Circuits

🔀 Parallel Circuits

What Makes a Parallel Circuit?

Key Properties of Parallel Circuits

Part 3: Combination Circuits

🧩 Combination Circuits

Circuit Reduction Strategy

Step-by-Step Method

Part 4: Voltage & Current Dividers

⚡ Voltage Dividers & Current Dividers

The Voltage Divider

The Formula

Part 5: Power in Circuits

💡 Power in Circuits

Power Equations

Part 6: Capacitors in Circuits

🔋 Capacitors in Series & Parallel

Part 7: Synthesis & AP Review

🎯 Synthesis & AP Review

Circuit Analysis Strategy

Step-by-Step Approach

Example

If One Breaks...

Why Is the Voltage the Same?

Total Resistance

Current Division in Parallel

Example

If One Branch Breaks...

Special Case: Identical Resistors in Parallel

Example

How to Tell Series from Parallel

Example: Three-Resistor Combination

Four-Resistor Combination

Step 1: Simplify the parallel pair

Step 2: Add all series resistances

Step 3: Total current

Step 4: Voltages and currents

Why This Works

Example

Key Insight

Potentiometers

How It Works

Applications

The Current Divider

The Formula

Key Insight — "Opposite" from Voltage Divider!

Example

Which Form to Use?

Units

Conservation of Energy

Power in Series Circuits

Example: Series

Power in Parallel Circuits

Example: Parallel

Series vs. Parallel Summary

Lightbulb Brightness & Power

Identical Bulbs (each resistance RRR)

Why?

Capacitors in Parallel

Parallel Capacitance Formula

Why?

Example

Capacitors in Series

Series Capacitance Formula

Key Fact

Why?

Example

Resistors vs. Capacitors: The Flip

Memory Trick

Energy Stored in a Capacitor

Identical Bulbs (each resistance $R$ )