1. Current and Potential Difference Rules in DC Circuits

Before combining resistors or calculating power, you need to know how current and potential difference (pd) behave in series and parallel arrangements. These rules follow directly from two conservation principles.

Current rules

In a series circuit, the current is the same through every component. This is because charge cannot accumulate at any point -- the charge entering a component each second must equal the charge leaving it each second.

At a junction (parallel circuit), the total current entering the junction equals the total current leaving it. This is a direct consequence of conservation of charge -- charge is neither created nor destroyed.

Potential difference rules

For components in series, the total pd across the combination equals the sum of the individual pds:

This follows from conservation of energy: each coulomb of charge delivers some energy to each component, and the sum of these energy transfers equals the total energy supplied per coulomb by the source.

For components in parallel, the pd across each branch is the same. Every coulomb of charge that passes through any one of the parallel branches transfers the same amount of energy, regardless of which branch it takes.

The next figure focuses on a series circuit; notice that the same current passes through every resistor while the individual voltmeter readings add to the battery emf.

[DIAGRAM: asset_name: 5.1.4 - Circuits - Diagram 1; asset_slug: 5.1.4 - Circuits - Diagram 1; recommended_method: retained_png; description: A series circuit with a battery of emf 12 V and three resistors. Voltmeters across the resistors read V_1, V_2, V_3 such that V_1 + V_2 + V_3 = 12 V. A single ammeter shows the same current I through each resistor.]

The following figure shows the parallel case; notice that both branches share the same pd while the battery current splits into branch currents that add to the total.

[DIAGRAM: asset_name: 5.1.4 - Circuits - Diagram 2; asset_slug: 5.1.4 - Circuits - Diagram 2; recommended_method: retained_png; description: A parallel circuit with a battery. Two branches, each containing a resistor. Voltmeters across each branch read the same value V. Ammeters in each branch read I_1 and I_2. The total current from the battery is I = I_1 + I_2.]

2. Kirchhoff's Laws

The current and pd rules described above are formalised as Kirchhoff's two laws.

Mathematically, at any junction:

For example, if three wires meet at a junction and currents of 0.5 A and 1.0 A flow into the junction along two of the wires, the current flowing out along the third wire must be 1.5 A.

Mathematically, for any loop:

Each time a charge carrier completes a loop, the energy it gains from sources of emf equals the energy it delivers to resistive components.

3. Combining Resistors

Resistors in series

When resistors are connected in series, the same current $I$ passes through each one. The total pd across the combination is:

Dividing through by $I$ gives the total resistance:

where $R_T$ is the total resistance, and $R_1, R_2, R_3, \dots$ are the individual resistances, all measured in ohms ($\Omega$). This relationship is worth learning fluently because it lets you simplify any series section quickly.

The total resistance in series is always greater than any individual resistance, because each resistor adds to the opposition to current flow.

Resistors in parallel

When resistors are connected in parallel, the pd $V$ across each is the same. The total current is:

Since $I = \frac{V}{R_T}$, dividing through by $V$ gives:

This relationship is worth remembering too. The total resistance of a parallel combination is always less than the smallest individual resistance, because providing additional paths for current reduces the overall opposition.

Useful shortcut for two resistors in parallel:

For example, a $3\,\Omega$ resistor in parallel with a $6\,\Omega$ resistor gives $R_T = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\,\Omega$.

Combining series and parallel in one circuit

Many circuits contain both series and parallel sections. The strategy is to simplify from the inside out: first calculate the equivalent resistance of any parallel combination, then add that result in series with the remaining resistors.

Worked example: A circuit contains a $10\,\Omega$ resistor and a $3\,\Omega$ resistor in series with a parallel combination of a $4\,\Omega$ resistor and a $16\,\Omega$ resistor. Find the total resistance.

Step 1 -- Find the resistance of the parallel combination:

Step 2 -- Add in series:

4. Cells in Series and Parallel

Cells (batteries) can also be combined in series or parallel, and the rules follow from the same conservation principles.

Cells in series

When cells are connected in series (positive terminal of one connected to the negative terminal of the next), the total emf is the sum of the individual emfs:

If a cell is connected in reverse (opposing the others), its emf is subtracted. For example, if a 3.0 V cell and a 2.0 V cell are connected in series but opposing each other, the net emf is $3.0 - 2.0 = 1.0\,\text{V}$.

The figure below shows the orientation of cells in series; notice that each positive terminal connects to the next negative terminal so the emfs add around the chain.

[DIAGRAM: asset_name: 5.1.4 - Circuits - Diagram 3; asset_slug: 5.1.4 - Circuits - Diagram 3; recommended_method: retained_png; description: Three cells connected in series, each labelled with emf epsilon_1, epsilon_2, epsilon_3. The positive terminal of each connects to the negative terminal of the next. Total emf = epsilon_1 + epsilon_2 + epsilon_3.]

Identical cells in parallel

When $n$ identical cells, each of emf $\varepsilon$, are connected in parallel, the total emf remains equal to the emf of a single cell:

This is because the current splits equally between the cells, so each cell only needs to supply a fraction $I/n$ of the total current. Each charge carrier passes through only one cell, gaining energy $\varepsilon$ per coulomb regardless of how many cells are in parallel. The advantage of connecting identical cells in parallel is that the effective internal resistance is reduced to $r/n$ (where $r$ is the internal resistance of one cell), allowing the combination to supply a larger current with less energy wasted internally.

5. Energy and Power Equations

Energy transferred

When a charge $Q$ moves through a potential difference $V$, the energy transferred is $E = QV$. Since current $I = Q/t$, we can write $Q = It$, giving:

where $E$ is energy in joules (J), $I$ is current in amperes (A), $V$ is potential difference in volts (V), and $t$ is time in seconds (s).

Electrical power

Power is the rate of energy transfer. Dividing $E = IVt$ by $t$:

where $P$ is power in watts (W).

Substituting $V = IR$ into $P = IV$ gives two alternative forms:

These three forms ($P = IV$, $P = I^2 R$, $P = V^2/R$) are all equivalent. Choose the form that uses the quantities you already know:

• Use $P = IV$ when you know current and pd.
• Use $P = I^2 R$ when you know current and resistance (useful for series circuits where current is the common quantity).
• Use $P = V^2 / R$ when you know pd and resistance (useful for parallel circuits where pd is the common quantity).

Resistance heating

The heating effect of current is caused by collisions between charge carriers (electrons) and the positive ions of the conducting material. Each collision transfers kinetic energy from the electron to the lattice, raising the temperature. The energy dissipated as heat per second is $P = I^2 R$.

Worked example: A lamp has a power rating of 60 W and is connected to a 240 V supply. Find the current through the lamp and the energy transferred in 2 minutes.

Current: $I = \frac{P}{V} = \frac{60}{240} = 0.25\,\text{A}$

Energy: $E = Pt = 60 \times 120 = 7200\,\text{J}$ (converting 2 minutes to 120 seconds).

Alternatively, $E = IVt = 0.25 \times 240 \times 120 = 7200\,\text{J}$.

Now let us bring together the resistor rules and the power equations in a more demanding problem.

6. Circuit Analysis Strategy

When tackling any DC circuit problem, follow a systematic approach:

1. Identify which components are in series and which are in parallel.
2. Simplify the circuit by replacing parallel combinations with their equivalent resistance, working from the innermost groups outward.
3. Calculate the total resistance of the simplified circuit.
4. Find the total current using $I = V / R_T$ (or $I = \varepsilon / R_T$ for the full circuit).
5. Work back outward to find the current through and pd across each individual component.
6. Calculate power for individual components using whichever form of the power equation is most convenient.

Always check your answer: the pds around any complete loop must sum to the emf, and the currents entering any junction must sum to the currents leaving it.

Once you can do the arithmetic, the last check is whether the current-sharing idea feels intuitive.