Series and Parallel Circuits

There are two fundamentally different ways to connect components in a circuit, and they behave so differently that learning the rules of each is one of the most useful things you can do in this topic. In a series circuit the components are joined end to end in a single loop; in a parallel circuit they are connected side by side on separate branches. The current, the potential difference and the total resistance follow completely different rules in the two arrangements — and knowing which rule applies lets you predict and calculate what every meter in a circuit will read. This lesson, part of Topic P3 (Electricity) of OCR Gateway Science A, sets out the series and parallel rules side by side and works through calculations comparing the two.

By the end of this lesson you should be able to describe series and parallel connections, state and apply the rules for current, potential difference and total resistance in each, explain why total resistance falls when resistors are added in parallel, and carry out worked calculations on both kinds of circuit.

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Series and Parallel Connections

In a series circuit, the components are connected one after another in a single loop, so there is only one path for the current to follow. The same current must pass through every component in turn — a string of old fairy lights wired in series is the classic example (and the reason that when one bulb failed, the whole string went out).

In a parallel circuit, the components are connected on separate branches between the same two points, so there is more than one path for the current. The current splits between the branches and recombines afterwards. The lights and sockets in your home are wired in parallel, so each can be switched on and off independently and each gets the full mains voltage.

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The Rules for a Series Circuit

In a series circuit there is one path, and three rules follow from that:

• Current is the same everywhere. Because there is only one path, the same current flows through every component: $I_1 = I_2 = I_3 = \ldots$I1​=I2​=I3​=… An ammeter placed anywhere in the loop reads the same value.
• Potential differences add up. The p.d. of the supply is shared between the components, so the individual p.d.s add up to the supply voltage: $V_{\text{total}} = V_1 + V_2 + V_3 + \ldots$Vtotal​=V1​+V2​+V3​+… A bigger resistor takes a bigger share of the voltage.
• Total resistance is the sum of the resistances. The resistances simply add: $R_{\text{total}} = R_1 + R_2 + R_3 + \ldots$Rtotal​=R1​+R2​+R3​+… Adding another resistor in series always increases the total resistance, because the charge has more opposition to push through.

Worked Example 1 — A series circuit

Two resistors, $4\ \Omega$4 Ω and $6\ \Omega$6 Ω, are connected in series with a $12 \text{ V}$12 V battery. Calculate the total resistance, the current in the circuit, and the potential difference across each resistor.

Step 1 — total resistance: $R_{\text{total}} = 4 + 6 = 10\ \Omega$Rtotal​=4+6=10 Ω.

Step 2 — current (same everywhere) from $I = \dfrac{V}{R}$I=RV​: $I = \dfrac{12}{10} = 1.2 \text{ A}$I=1012​=1.2 A.

Step 3 — p.d. across the $4\ \Omega$4 Ω resistor: $V = IR = 1.2 \times 4 = 4.8 \text{ V}$V=IR=1.2×4=4.8 V.

Step 4 — p.d. across the $6\ \Omega$6 Ω resistor: $V = IR = 1.2 \times 6 = 7.2 \text{ V}$V=IR=1.2×6=7.2 V.

Check: the two p.d.s add to $4.8 + 7.2 = 12 \text{ V}$4.8+7.2=12 V, the supply voltage — as they must in series.

Exam Tip: In series, current is the same everywhere and the voltages add up to the supply. A good check on any series calculation is that your component p.d.s sum to the battery voltage.

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The Rules for a Parallel Circuit

In a parallel circuit there is more than one path, and the rules are effectively the reverse of the series ones:

• Potential difference is the same across each branch. Each branch is connected across the same two points, so each gets the full supply voltage: $V_1 = V_2 = V_3 = \ldots = V_{\text{supply}}$V1​=V2​=V3​=…=Vsupply​. This is why every light in your house runs at the full mains voltage.
• Currents add up. The current splits between the branches and the branch currents add up to the total current drawn from the supply: $I_{\text{total}} = I_1 + I_2 + I_3 + \ldots$Itotal​=I1​+I2​+I3​+… A branch with a smaller resistance takes a larger share of the current.
• Total resistance is LESS than the smallest individual resistance. Adding another resistor in parallel lowers the total resistance, because it provides an extra path for the current to flow through.

Worked Example 2 — A parallel circuit

Two resistors, $4\ \Omega$4 Ω and $6\ \Omega$6 Ω, are connected in parallel across a $12 \text{ V}$12 V battery. Calculate the current in each branch and the total current from the battery.

Step 1 — each branch has the full $12 \text{ V}$12 V across it.

Step 2 — current in the $4\ \Omega$4 Ω branch: $I = \dfrac{V}{R} = \dfrac{12}{4} = 3 \text{ A}$I=RV​=412​=3 A.

Step 3 — current in the $6\ \Omega$6 Ω branch: $I = \dfrac{V}{R} = \dfrac{12}{6} = 2 \text{ A}$I=RV​=612​=2 A.

Step 4 — total current = sum of branch currents: $I_{\text{total}} = 3 + 2 = 5 \text{ A}$Itotal​=3+2=5 A.

Answer: $3 \text{ A}$3 A in the $4\ \Omega$4 Ω branch, $2 \text{ A}$2 A in the $6\ \Omega$6 Ω branch, and $5 \text{ A}$5 A in total. Notice the smaller resistor ($4\ \Omega$4 Ω) carries the larger current.

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Why Total Resistance Falls in Parallel

It can seem strange that adding a resistor can reduce the total resistance, so it is worth understanding why. When you connect a second resistor in parallel, you give the current an extra path to flow through. With more paths available, more current can flow for the same supply voltage. Since the total resistance is the supply voltage divided by the total current ($R_{\text{total}} = V/I_{\text{total}}$Rtotal​=V/Itotal​), a larger total current means a smaller total resistance.

An everyday analogy: opening a second checkout in a supermarket lets more shoppers through per minute, even though each till is no faster — the combined "resistance" to the flow of shoppers has gone down. In the worked example above, the total current was $5 \text{ A}$5 A from a $12 \text{ V}$12 V supply, so the total resistance is $R_{\text{total}} = \dfrac{12}{5} = 2.4\ \Omega$Rtotal​=512​=2.4 Ω — which is less than the smaller of the two resistors ($4\ \Omega$4 Ω), exactly as the rule predicts.