GCSE · Physics · Edexcel · Spec 1PH0

Calculations in series circuits

Two bulbs, one loop, one cell. Does the first bulb use up the current? Does each get the cell's whole push? Neither — and the reason makes series sums easy.

Who gets how much of the push?

Resistance of R₁: 10 Ω. p.d. across R₂: 4.0 V

At every setting the current is the same through R₁ and R₂, and the two p.d.s add up to the 6 V of the cell.

Drag the handle to change the resistance of R₁. R₂ stays at 20 Ω and the cell stays at 6 V. Read the p.d. and current beside each resistor.

Series circuit · voltmeters

Where does each voltmeter sit?

Showing 1 layer: Circuit

Explore

tap a voltmeter ↓

Check your gut

Which of these sounds like you?

A cell is connected to two resistors in one single loop. Think about the current, and about the cell's p.d.

Which is closest to what you think right now?

The current gets used up as it passes through each resistor, so there's less left after the first one.Each resistor gets the whole p.d. of the cell.The current is the same all the way round the loop, and the p.d.s across the resistors add up to the cell's p.d.The cell's p.d. is split equally between the resistors, whatever their resistances.

How sure are you?

I'm sureI think soI'm guessing

The current gets used up as it passes through each resistor, so there's less left after the first one.

This is a common misconception

It's an easy idea to fall into, because a resistor feels like it 'takes something' from the loop. But current is the charge flowing past a point each second, and in a series circuit it is the same everywhere. The same current goes through the first resistor and the second. What gets shared out is the p.d., not the current.

Each resistor gets the whole p.d. of the cell.

This is a common misconception

Test it with numbers. If both resistors had the full 6 V of a 6 V cell, you'd have 12 V across the components from a 6 V supply. Instead the p.d. of the supply is shared: the p.d.s across the components add up to the p.d. of the supply, so a 6 V cell might give 2 V and 4 V.

The current is the same all the way round the loop, and the p.d.s across the resistors add up to the cell's p.d.

This matches the evidence

Yes. Hold on to those two rules: one for current (the same everywhere) and one for p.d. (shared out, and the shares add up to the supply).

The cell's p.d. is split equally between the resistors, whatever their resistances.

This is a common misconception

Equal p.d.s need equal resistances. The p.d. across a component is determined by its resistance: a higher resistance needs a higher p.d. to push the same current through it, so it has the largest p.d. In the circuit at the top, R₁ and R₂ only shared the 6 V equally when both were 20 Ω.

Current: the same all the way round. P.d.: shared out, the biggest resistance takes the biggest share, and the shares add up to the supply.

The V = I × R triangle

Cover the quantity you want. What's left tells you what to do.

Tap the quantity you want to find. The triangle shows you the formula.

Cover potential difference, current or resistance to reveal its rearranged formula, then plug in numbers to solve.

Your turn

Two rules, then I = V ÷ R

A 9.0 V cell is connected in series with resistor A and resistor B. The p.d. across A is 6.0 V and B has a resistance of 12 Ω. Find the resistance of A.

Write down what we know and what we want: supply 9.0 V, p.d. across A 6.0 V, resistance of B 12 Ω. We want the resistance of A.
missing step

Which line is step 2?

The p.d. across B = 9.0 + 6.0 = 15 V.The p.d.s add up to the supply p.d., so the p.d. across B = 9.0 − 6.0 = 3.0 V.The p.d. across B = 9.0 V, because every component gets the cell's whole p.d.

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

I = V ÷ R

Same current everywhere. Shared p.d. that adds up to the supply.

What you need to know

A series circuit is a single loop. Current — the charge flowing past a given point in one second — is the same everywhere in it.
A cell's potential difference (p.d.) is the push its electric field gives to charges: the greater the p.d., the greater the force on the electrons.
Adding anything with resistance lowers the current; adding another cell in series raises the p.d. and the current.
Have a goSam adds a second resistor into a series loop and says, 'More stuff in the loop, so more current.' Is Sam right?
No. The current goes down.
A resistor adds resistance, so the same push from the cell now has to drive current through more resistance, and the current decreases.
The p.d.s across the components add up to the p.d. of the supply, so the cell's push is shared out.
A higher resistance needs a higher p.d. to push the same current through it, so that component has the largest p.d.
Have a goTwo resistors in series: one is 10 Ω and one is 30 Ω. Which has the larger p.d. across it?
The 30 Ω resistor.
The same current goes through both, and the larger resistance needs the larger p.d. to push it, so it takes the biggest share.
A voltmeter across a component reads that component's p.d.; across the cell it reads the sum of them all.
For any component, I = V ÷ R, with current in amperes (A), p.d. in volts (V) and resistance in ohms (Ω).
Rearranged, V = I × R and R = V ÷ I: know any two of the three and you can calculate the third.
Have a goA 4 Ω resistor has a current of 2 A in it. Which arrangement of I = V ÷ R finds the p.d. across it, and what is it?
V = I × R = 2 × 4 = 8 V.
Multiplying both sides of I = V ÷ R by R leaves V on its own, and p.d. is measured in volts.
Method: write the given quantities and the one to find, write the arrangement, substitute, calculate, add the unit.
If only one value in I = V ÷ R is known, use the current or p.d. rule to find a second.

The big picture

In a series circuit the current is the same everywhere, and the cell's p.d. is shared between the components so the shares add up to the supply. Add I = V ÷ R and its rearrangements and you can find any missing current, p.d. or resistance.

Key points

1A series circuit is one loop: the current is the same everywhere.

2The p.d.s across the components add up to the supply p.d.

3The component with the largest resistance has the largest p.d.

4I = V ÷ R, V = I × R, R = V ÷ I — know two, find the third, include the unit.

.we-_R_18slubsnpfiv5ubrb_ [data-we-step]{display:flex !important}.we-_R_18slubsnpfiv5ubrb_ [data-we-solution]{display:block !important}.we-_R_18slubsnpfiv5ubrb_ [data-we-reveal]{display:none !important}

Worked example

Problem

A bulb has a p.d. of 12 V across it and a resistance of 40 Ω. What is the current through the bulb?

⚠ Watch out

Giving every component the cell's full p.d., or treating current as used up. The current is the same all the way round; the p.d. is what's shared and adds up to the supply. In I = V ÷ R, use the V and R of the same component.

🧠

Memory hook

Same current, shared p.d. The current is the same all the way round; the cell's push is shared out and the shares add up to the supply.

✓

Check yourself

A cell supplies 10 V to two resistors in series. One has 7 V across it. What is across the other, and which resistor has the larger resistance?

Flashcards

(14)

What is a series circuit?

An electrical circuit with a single loop.

What is current?

The amount of charge flowing past a given point in one second.

How does the current compare at different points in a series circuit?

It is the same everywhere in the circuit.

What does the p.d. of a cell measure?

The push its electric field gives to charges. The greater the p.d., the greater the force on the electrons.

What happens to the current when you add anything with resistance to a series circuit?

The total resistance increases and the current decreases. The same p.d. gives the same push, but there is more resistance.

What happens when you add another cell in series?

The p.d. increases, so the push on the electrons and the current increase. Two 1.5 V cells in series give 3.0 V.

How do the p.d.s across the components compare with the supply p.d.?

They add up to it. The supply p.d. is shared between the components.

Which component in a series circuit has the largest p.d. across it?

The one with the highest resistance. It needs the highest p.d. to push the same current through it.

What does a voltmeter across a component measure?

The p.d. across that component.

What does a voltmeter across the cell read?

The same as the sum of the p.d.s across all the components in the single loop.

Write the equation linking current, p.d. and resistance, with units.

I = V ÷ R. Current in amperes (A), p.d. in volts (V), resistance in ohms (Ω).

How do you rearrange I = V ÷ R?

V = I × R, then R = V ÷ I. Any two known values give the third.

What are the steps of the calculation method?

Write the given quantities and the one to find, write the arrangement, substitute, calculate, and write the answer with its unit.

Only one value in I = V ÷ R is known. What do you do?

Apply the series rule for current or p.d. to find a second value.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

Learning with Lightbulb is opening soon

You can use this lesson now. Join the waitlist and we'll let you know when the full Lightbulb experience is ready.

Keep me posted

Who gets how much of the push?

Which of these sounds like you?

Two rules, then I = V ÷ R

More Edexcel GCSE Physics topics