DC Circuits

DC Circuits

Info: Board Coverage AQA Paper 1 | Edexcel CP2 | OCR (A) Paper 2 | CIE P2

Explore the simulation above to develop intuition for this topic.

Intuition

Circuits are plumbing for electrons: Think of a battery as a pump, wires as pipes, and components (resistors, LEDs) as valves or restrictions. Current flows from high voltage to low voltage, just as water flows from high pressure to low pressure. Kirchhoff’s laws are just conservation of charge (no electrons accumulate at junctions) and conservation of energy (voltage around any loop is zero).

Why it matters: DC circuits are the foundation of all electronics — from simple flashlight circuits to complex computer processors. Understanding Kirchhoff’s laws lets you analyse any circuit, no matter how complicated, by breaking it into manageable loops and junctions.

The key insight: Kirchhoff’s voltage law (KVL) is just conservation of energy applied to circuits. The battery provides energy, and components consume it. The total energy provided must equal the total energy consumed — that’s why voltage around any loop sums to zero. This is the same principle as a roller coaster: what goes up must come down.

1. Kirchhoff’s Laws

Kirchhoff’s First Law (Current Law)

Definition. Kirchhoff’s first law states that the algebraic sum of currents entering any Junction in an electrical circuit is zero, reflecting the conservation of electric charge: $\sum I = 0$.

The sum of currents entering a junction equals the sum of currents leaving it.

$\sum I_{\mathrm{in}} = \sum I_{\mathrm{out}}$

Derivation from charge conservation. Charge cannot accumulate at a junction (in steady state). The rate of charge flowing in must equal the rate flowing out:

$\frac{dQ_{\mathrm{in}}}{dt} = \frac{dQ_{\mathrm{out}}}{dt} \implies \sum I_{\mathrm{in}} = \sum I_{\mathrm{out}}$

$\square$

Kirchhoff’s Second Law (Voltage Law)

Definition. Kirchhoff’s second law states that the algebraic sum of potential differences around Any closed loop in an electrical circuit is zero, reflecting the conservation of energy: $\sum \mathcal{E} = \sum V$.

The sum of e.m.f.s around any closed loop equals the sum of p.d.s around that loop.

$\sum \mathcal{E} = \sum V$

Derivation from energy conservation. As a unit charge moves around a closed loop, the total Energy gained from sources must equal the total energy lost to components. Otherwise, energy would Be created or destroyed:

$\frac{W_{\mathrm{gained}}}{Q} = \frac{W_{\mathrm{lost}}}{Q} \implies \sum \mathcal{E} = \sum IR$

$\square$

2. Series and Parallel Resistors

Series Combination

For resistors $R_1, R_2, \ldots, R_n$ in series, the same current $I$ flows through each.

By Kirchhoff’s second law:

$V = V_1 + V_2 + \cdots + V_n = IR_1 + IR_2 + \cdots + IR_n$

$V = I(R_1 + R_2 + \cdots + R_n)$

The equivalent resistance is:

$\boxed{R_{\mathrm{series}} = R_1 + R_2 + \cdots + R_n}$

Parallel Combination

For resistors in parallel, the same p.d. $V$ is across each.

By Kirchhoff’s first law:

$I = I_1 + I_2 + \cdots + I_n = \frac{V}{R_1} + \frac{V}{R_2} + \cdots + \frac{V}{R_n}$

$I = V\left(\frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}\right)$

The equivalent resistance satisfies $V/R_{\mathrm{parallel}} = I$:

$\boxed{\frac{1}{R_{\mathrm{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}}$

For two resistors in parallel:

$R_{\mathrm{parallel}} = \frac{R_1 R_2}{R_1 + R_2}$

Proof. $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{R_1 + R_2}{R_1 R_2}$Hence $R = \frac{R_1 R_2}{R_1 + R_2}$. $\square$

Intuition. Adding resistors in series always increases total resistance (current must pass Through all of them). Adding resistors in parallel always decreases total resistance (current has More paths to flow through). The parallel equivalent is always less than the smallest individual Resistor.

3. Potential Dividers

Definition. A potential divider is a circuit consisting of two or more resistors connected in Series across a voltage supply, which produces an output voltage that is a fraction of the input Voltage, determined by the ratio of resistances.

A potential divider consists of two or more resistors in series across a voltage supply. The Voltage is divided in proportion to the resistances.

For two resistors $R_1$ and $R_2$ in series across supply $V_{\mathrm{in}}$:

$V_{\mathrm{out}} = V_{\mathrm{in}} \times \frac{R_2}{R_1 + R_2}$

Derivation. The current through both resistors is $I = V_{\mathrm{in}}/(R_1 + R_2)$. The p.d. Across $R_2$ is:

$V_{\mathrm{out}} = IR_2 = \frac{V_{\mathrm{in}}}{R_1 + R_2} \cdot R_2 = V_{\mathrm{in}} \cdot \frac{R_2}{R_1 + R_2}$

$\square$

Potentiometer

A potentiometer is a continuous potential divider — a single resistive track with a sliding Contact. It can provide any output voltage from 0 to $V_{\mathrm{in}}$.

Potential Divider with a Sensor

A thermistor or LDR can replace one resistor, creating a voltage that varies with temperature or Light intensity:

$V_{\mathrm{out}} = V_{\mathrm{in}} \cdot \frac{R_{\mathrm{sensor}}}{R_{\mathrm{fixed}} + R_{\mathrm{sensor}}}$

4. Internal Resistance

Definition. Internal resistance $r$ is the resistance within a power source that opposes the Flow of current and causes energy to be dissipated within the source, reducing the terminal Potential difference below the e.m.f.

Definition. Terminal potential difference $V$ is the potential difference across the terminals Of a power source when it is delivering current, equal to the e.m.f. Minus the potential difference Across the internal resistance: $V = \mathcal{E} - Ir$.

Real power sources have internal resistance $r$Which causes the terminal p.d. To drop as Current increases.

Derivation of $V = \mathcal{E} - Ir$

Consider a cell of e.m.f. $\mathcal{E}$ and internal resistance $r$ connected to an external circuit Of resistance $R$. The total resistance in the circuit is $R + r$.

$I = \frac{\mathcal{E}}{R + r}$

The terminal p.d. (voltage across the external circuit) is:

$V = IR = \frac{\mathcal{E}R}{R + r} = \mathcal{E}\left(\frac{R}{R+r}\right)$

Alternatively, by Kirchhoff’s second law around the full circuit:

$\mathcal{E} = V + Ir \implies \boxed{V = \mathcal{E} - Ir}$

The “lost volts” is $Ir$ — the p.d. Across the internal resistance.

Maximum power transfer. The power delivered to the load is:

$P = I^2R = \frac{\mathcal{E}^2 R}{(R+r)^2}$

Differentiating with respect to $R$ and setting $dP/dR = 0$ gives $R = r$. Maximum power is Delivered when the load resistance equals the internal resistance.

Measuring Internal Resistance

Method 1: Vary the load. Measure $V$ and $I$ for several different values of $R$. Plot $V$ Against $I$.

$V = \mathcal{E} - Ir$

This is a straight line with gradient $= -r$ and $y$-intercept $= \mathcal{E}$.

Method 2: Direct calculation. With two measurements $(V_1, I_1)$ and $(V_2, I_2)$:

$r = \frac{V_1 - V_2}{I_2 - I_1}$

5. Thevenin’s Theorem

Thevenin’s Theorem. Any linear circuit with two terminals can be replaced by an equivalent Circuit consisting of:

• A single voltage source $V_{\mathrm{Th}}$ (the Thevenin voltage)
• A single series resistance $R_{\mathrm{Th}}$ (the Thevenin resistance)

Finding $V_{\mathrm{Th}}$

The Thevenin voltage is the open-circuit voltage between the two terminals (i.e., the voltage when Nothing is connected across them).

Finding $R_{\mathrm{Th}}$

The Thevenin resistance is the resistance between the two terminals when all independent voltage Sources are replaced by short circuits (and current sources by open circuits).

Intuition. Thevenin’s theorem is powerful because it lets you simplify any complex circuit into A simple source-resistor pair. Once you know $V_{\mathrm{Th}}$ and $R_{\mathrm{Th}}$You can Predict the behaviour of the circuit for any load.

6. Wheatstone Bridge

A Wheatstone bridge is a circuit arrangement of four resistors in a diamond configuration with a Galvanometer connected between the midpoint nodes. It provides a precise method for measuring Unknown resistances.

Balance Condition

Consider four resistors $R_1, R_2, R_3, R_4$ arranged so that $R_1$ and $R_2$ form one branch, $R_3$ And $R_4$ form another, with a galvanometer of resistance $R_g$ connected between the junction of $R_1$-$R_2$ and the junction of $R_3$-$R_4$.

Derivation. At balance, no current flows through the galvanometer ($I_g = 0$).

By KCL at each junction, the current through $R_1$ equals the current through $R_2$ (call it $I_1$), And the current through $R_3$ equals the current through $R_4$ (call it $I_2$).

Since $I_g = 0$The p.d. Across the galvanometer is zero, so:

$V_A = V_B$

Where $A$ is the junction of $R_1$ and $R_3$And $B$ is the junction of $R_2$ and $R_4$.

The p.d. Across $R_1$: $V_{R_1} = I_1 R_1$. The p.d. Across $R_3$: $V_{R_3} = I_2 R_3$.

For the galvanometer p.d. To be zero:

$I_1 R_1 = I_2 R_3 \quad \mathrm{...(i)}$

Similarly, considering the other path:

$I_1 R_2 = I_2 R_4 \quad \mathrm{...(ii)}$

Dividing (i) by (ii):

$\frac{R_1}{R_2} = \frac{R_3}{R_4}$

$\boxed{R_1 R_4 = R_2 R_3}$

$\square$

Meter Bridge

A meter bridge is a practical form of the Wheatstone bridge. A uniform resistance wire of length 1 m replaces two of the resistors. The balance point is found by sliding a contact along the wire.

At balance: $\frac{R}{S} = \frac{l_1}{l_2} = \frac{l}{100 - l}$

Where $R$ is the known resistance, $S$ is the unknown, and $l$ is the balance length.

7. Norton’s Theorem

Norton’s Theorem. Any linear two-terminal circuit can be replaced by an equivalent circuit Consisting of a current source $I_N$ in parallel with a resistance $R_N$.

Finding $I_N$

The Norton current is the short-circuit current between the two terminals (the current that flows When the terminals are directly connected).

Finding $R_N$

The Norton resistance is the same as the Thevenin resistance: the resistance between the two Terminals when all independent sources are replaced by their internal resistances (voltage sources Short-circuited, current sources open-circuited).

Source Transformation

Thevenin and Norton equivalents are related by:

$I_N = \frac{V_{\mathrm{Th}}}{R_{\mathrm{Th}}}, \qquad R_N = R_{\mathrm{Th}}$

A voltage source $V_{\mathrm{Th}}$ in series with $R_{\mathrm{Th}}$ is equivalent to a current Source $I_N$ in parallel with $R_N$.

Proof. For the Thevenin equivalent: open-circuit voltage $= V_{\mathrm{Th}}$Short-circuit Current $= V_{\mathrm{Th}}/R_{\mathrm{Th}}$. For the Norton equivalent: short-circuit current $= I_N$Open-circuit voltage $= I_N R_N$. For equivalence: $I_N = V_{\mathrm{Th}}/R_{\mathrm{Th}}$ And $R_N = R_{\mathrm{Th}}$. $\square$

8. Multi-Loop Circuit Analysis

For circuits with multiple loops, a systematic approach is needed.

Loop Current Method (Mesh Analysis)

1. Assign a clockwise loop current to each independent loop.
2. Apply KVL around each loop, expressing voltage drops in terms of loop currents.
3. Solve the resulting system of simultaneous equations.

Worked example. Consider a circuit with two loops. Loop 1 contains a 12 V battery and $R_1 = 4\,\Omega$. Loop 2 contains a 6 V battery and $R_2 = 2\,\Omega$. The shared branch contains $R_3 = 6\,\Omega$.

Assign loop currents $I_1$ (loop 1, clockwise) and $I_2$ (loop 2, clockwise).

Loop 1 (KVL): $12 = 4I_1 + 6(I_1 - I_2)$

Loop 2 (KVL): $6 = 2I_2 + 6(I_2 - I_1)$

Expanding:

$12 = 10I_1 - 6I_2 \quad \mathrm{...(i)}$ $6 = -6I_1 + 8I_2 \quad \mathrm{...(ii)}$

From (i): $I_2 = (10I_1 - 12)/6$.

Substituting into (ii): $6 = -6I_1 + 8(10I_1 - 12)/6 = -6I_1 + (80I_1 - 96)/6$.

$36 = -36I_1 + 80I_1 - 96$. $132 = 44I_1$. $I_1 = 3.0$ A.

$I_2 = (30 - 12)/6 = 3.0$ A.

Current through $R_3$: $I_1 - I_2 = 0$ A (the shared resistor carries no current in this case).

Number of Independent Equations

For a circuit with $b$ branches and $n$ nodes:

• KCL provides $n - 1$ independent equations.
• KVL provides $b - n + 1$ independent equations.
• Total: $(n - 1) + (b - n + 1) = b$ equations for $b$ unknown branch currents.

9. Measuring Instruments and Loading Effects

Ideal vs Real Instruments

An ideal voltmeter has infinite resistance ($R_V \to \infty$) and draws zero current. A real Voltmeter has finite resistance ( $10\,\mathrm k\Omega$ to $10\,\mathrm M\Omega$ for Digital meters).

An ideal ammeter has zero resistance ($R_A \to 0$) and drops zero voltage. A real ammeter Has finite (but small) resistance.

Loading Error

When a voltmeter of resistance $R_V$ is connected across a resistor $R_2$ in a potential divider, The parallel combination changes the circuit:

$R_{\mathrm{eff}} = \frac{R_2 R_V}{R_2 + R_V}$

The measured voltage is:

$V_{\mathrm{measured}} = V_{\mathrm{in}} \cdot \frac{R_{\mathrm{eff}}}{R_1 + R_{\mathrm{eff}}}$

Which is always less than the true voltage $V_{\mathrm{true}} = V_{\mathrm{in}} \cdot R_2/(R_1 + R_2)$.

Proof that loading always reduces the reading. Since $R_{\mathrm{eff}} \lt R_2$ (parallel Resistance is less than the smallest), and $V_{\mathrm{out}}$ is an increasing function of the lower Resistance in the divider, we have $V_{\mathrm{measured}} \lt V_{\mathrm{true}}$. $\square$

Voltmeter Multiplier

To extend the range of a galvanometer (full-scale current $I_g$Internal resistance $R_g$) to Measure voltage up to $V_{\mathrm{range}}$:

A series “multiplier” resistor $R_s$ is added so that at full scale:

$V_{\mathrm{range}} = I_g(R_g + R_s)$

$\boxed{R_s = \frac{V_{\mathrm{range}}}{I_g} - R_g}$

Ammeter Shunt

To extend the range of a galvanometer to measure current up to $I_{\mathrm{range}}$:

A parallel “shunt” resistor $R_{\mathrm{shunt}}$ diverts excess current so that $I_g$ flows through The galvanometer:

$I_g R_g = (I_{\mathrm{range}} - I_g) R_{\mathrm{shunt}}$

$\boxed{R_{\mathrm{shunt}} = \frac{R_g \cdot I_g}{I_{\mathrm{range}} - I_g}}$

Problem Set

Common Pitfalls

1. Confusing EMF and potential difference. EMF is the total energy per unit charge supplied; PD is the energy per unit charge transferred to a component.

2. Forgetting that ammeters are connected in series and voltmeters in parallel.

3. Incorrectly applying Kirchhoff’s second law by missing components in a loop.

4. Forgetting to include units in final answers, especially when working with derived units like $\text{N}\,\text{kg}^{-1}\,\text{m}^2$.

5. Neglecting air resistance or assuming ideal conditions when the question specifies a real-world scenario.

6. Rounding intermediate answers too early, which compounds errors in multi-step calculations.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Cross-References

• Capacitance — Capacitors store charge and are analysed using similar circuit techniques to resistors.
• Quantum Physics — The photoelectric effect relies on circuit measurements to determine photon energy.
• Fields — Electric fields underpin the potential differences that drive current in DC circuits.
• Measurements and Uncertainties — Instrument loading errors and precision requirements connect directly to measurement theory.