🔗 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

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 $R_k$ in a series circuit is:

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

Example

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

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

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

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

If One Breaks...

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!

Series Circuit Calculation Drill

A 24 V battery is connected to three resistors in series: $R_1 = 4 \; \Omega$, $R_2 = 8 \; \Omega$, $R_3 = 12 \; \Omega$.

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

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.

1. Voltage across the second resistor (in V)
2. Resistance of the first resistor (in $\Omega$)
3. Resistance of the second resistor (in $\Omega$)

Exit Quiz — Series 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

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.

Total Resistance

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 $12$ 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: $R_{\text{total}} = \frac{4 \times 12}{4 + 12} = \frac{48}{16} = 3 \; \Omega$. $I = 12/3 = 4$ 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: $R_1 = 10 \; \Omega$, $R_2 = 15 \; \Omega$, $R_3 = 30 \; \Omega$.

1. Current through $R_1$ (in A)
2. Current through $R_2$ (in A)
3. Current through $R_3$ (in A)
4. Total current from the battery (in A)
5. Total resistance of the circuit (in $\Omega$)

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!

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

🧩 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

1. Identify groups of resistors that are purely in series or purely in parallel
2. Replace each group with a single equivalent resistor
3. Repeat until you have a single equivalent resistance
4. Find total current using $I = V/R_{\text{total}}$
5. 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 $R_2$ and $R_3$)
• $I_2 = V_{23}/R_2 = 8/4 = 2$ 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.

1. Equivalent resistance of $R_2$ and $R_3$ in parallel (in $\Omega$)
2. Total circuit resistance (in $\Omega$)
3. Total current from the battery (in A)
4. Voltage across $R_1$ (in V)
5. Voltage across the parallel group (in V)
6. Current through $R_2$ (in A)

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
• $V_4 = 2.12 \times 3 \approx 6.35$ V
• $I_1 = 3.18/2 \approx 1.59$ A
• $I_2 = 3.18/6 \approx 0.53$ A

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

Exit Quiz — Combination Circuits

⚡ 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 $V_{\text{in}}$:

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

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

Why This Works

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

Example

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

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

Key Insight

• 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 $R_1$)
• If $R_1 = R_2$: $V_{\text{out}} = V_{\text{in}}/2$ (voltage splits equally)

Voltage Divider Drill

1. 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)
2. 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 $R_1$ be? (in k$\Omega$)
3. 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 larger share of the current.

Example

$I_{\text{total}} = 6$ A splits between $R_1 = 4 \; \Omega$ and $R_2 = 12 \; \Omega$ 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.

1. Current through $R_1$ (in A)
2. Current through $R_2$ (in A)
3. Voltage across the parallel combination (in V)

Exit Quiz — Dividers

💡 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:

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

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

Which Form to Use?

Units

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

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, $V = 16$ V:

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

$P_1 = (2)^2 \times 2 = 8 \text{ W}$ $P_2 = (2)^2 \times 6 = 24 \text{ W}$ $P_{\text{total}} = 8 + 24 = 32 \text{ W}$

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, $V = 12$ 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: $R_{\text{total}} = 3 \times 6/(3+6) = 2 \; \Omega$. $I = 12/2 = 6$ A. $P = 6 \times 12 = 72$ W ✓

Series vs. Parallel Summary

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 $R_3 = 12 \; \Omega$.

1. Total resistance of the circuit (in $\Omega$)
2. Total current (in A)
3. Power dissipated by $R_1$ (in W)
4. Power dissipated by $R_2$ (in W)
5. 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

🔋 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: $U = \frac{1}{2}CV^2$
• 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

$C_1 = 2 \; \mu\text{F}$, $C_2 = 3 \; \mu\text{F}$, $C_3 = 5 \; \mu\text{F}$ 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

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.

1. Total capacitance (in $\mu$F)
2. Charge stored on each capacitor (in $\mu$C)
3. Voltage across $C_1$ (in V)
4. Voltage across $C_2$ (in V)
5. Total energy stored (in $\mu$J)

Exit Quiz — Capacitors

🎯 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

1. Draw and label — Redraw the circuit neatly. Label all resistors/capacitors and the source.
2. Identify topology — Find series and parallel groups.
3. Simplify — Reduce combination groups one step at a time.
4. Solve for totals — Find $R_{\text{total}}$ (or $C_{\text{total}}$), then $I_{\text{total}}$ or $Q_{\text{total}}$.
5. Expand backward — Work back through each simplification step, applying:
   • Series: same $I$, voltages add
   • Parallel: same $V$, currents add
6. Verify — Check that voltages around any loop sum to zero, currents at any node balance, and total power balances.

Quick Reference

Common AP Mistakes

❌ Mistake 1: Confusing Resistor and Capacitor Rules

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!

Mistake Spotter

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:

1. Analyze the circuit before the change
2. Analyze the circuit after the change
3. 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 $R_3 = 24 \; \Omega$.

1. $R_{\text{total}}$ of the circuit (in $\Omega$)
2. Total current from the battery (in A)
3. Power dissipated by $R_1$ (in W)
4. Current through $R_3$ (in A)
5. Total power delivered by the battery (in W)

Round all answers to 3 significant figures.

Mastery Quiz — Series & Parallel Circuits