6th Sep '25

Circuits

The potential difference provided by cells connected in series is the sum of the potential difference of each cell

V_T = V₁ + V₂ + V₃

So if each cell had a potential difference of 1.5V across it, the battery would have a potential of V across it.

If one of the cells was placed the wrong way round then we have to take it away:

So if each cell had a potential difference of 1.5V across it, the battery would have a potential of V across it.

Although large circuits can become quite complicated, there are simple rules that can be used to analyse them. We often use shorthand to refer to measurements in the circuit:

Let:

R₁ = The of component 1
(Also used to refer to the physical component - Resistor 1)
V₁ = The across component 1
i₁ = The flowing through component 1
A₁ = The reading on the measuring the current flowing through component 1
Therefore: A₁ = i₁

While this is a logical way of shorthanding the measurements in a circuit, in the exam the question will 'invent' extra currents (ammeters) or potential differences (voltmeters) to test if you can work them out.
Make sure you study the circuit diagram in the questions for the specific subscript numbers

For components connected in series:

there is the current through each component
the total potential difference of the supply is between the components.
the total resistance is the of the resistance of each component

Therefore:

Let Current through R₁ = i₁ etc
Reading on Ammeter = i₁ = i₂ = i₃
V_T = V₁ + V₂ + V₃

We can replace all the resistors in the circuit with one equilavent resistor, with a resistance equal to the total resistance of the others. This is calculated using the following:

R_T = R₁ + R₂ + R₃

For components connected in parallel:

the potential difference across each component is the
the total current through the whole circuit is the of the currents through the separate components ie the current is between the branches.

Therefore:

V_T = V₁ = V₂ = V₃
A_T = A₁ + A₂ + A₃
Or
i_T = i₁ + i₂ + i₃
The greater the resistance the the current through that branch - in proportion. eg if one branch has twice the resistance of another, it will have the current flowing through it
The more parallel branches there are the the overall resistance, and hence a current will be drawn from the battery.

Build your own circuit:

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Let: R₁ = The of component 1 (Also used to refer to the physical component - Resistor 1) V₁ = The across component 1 i₁ = The flowing through component 1 A₁ = The reading on the measuring the current flowing through component 1 Therefore: A₁ = i₁
While this is a logical way of shorthanding the measurements in a circuit, in the exam the question will 'invent' extra currents (ammeters) or potential differences (voltmeters) to test if you can work them out. Make sure you study the circuit diagram in the questions for the specific subscript numbers
For components connected in series: there is the current through each component the total potential difference of the supply is between the components. the total resistance is the of the resistance of each component
Therefore: Let Current through R₁ = i₁ etc Reading on Ammeter = i₁ = i₂ = i₃ V_T = V₁ + V₂ + V₃
We can replace all the resistors in the circuit with one equilavent resistor, with a resistance equal to the total resistance of the others. This is calculated using the following: R_T = R₁ + R₂ + R₃
For components connected in parallel: the potential difference across each component is the the total current through the whole circuit is the of the currents through the separate components ie the current is between the branches.
Therefore: V_T = V₁ = V₂ = V₃ A_T = A₁ + A₂ + A₃ Or i_T = i₁ + i₂ + i₃ The greater the resistance the the current through that branch - in proportion. eg if one branch has twice the resistance of another, it will have the current flowing through it The more parallel branches there are the the overall resistance, and hence a current will be drawn from the battery.