A cell makes one end of the circuit positive and the other negative. This sets up a?potential difference?V?across the circuit
The potential difference across a component in a circuit is defined as the?energy transferred per unit charge?flowing from one point to another
The energy transfer is from electrical energy into other forms
Potential difference is measured in?volts (V). This is the same as a?Joule per coulomb (J C-1)
If a bulb has a voltage of 3 V, every coulomb of charge passing through the bulb will transfer 3 J of energy
The potential difference of a power supply connected in series is always shared between all the components in the circuit
The potential difference is the voltage across each component in a circuit
Calculating Potential Difference
The potential difference is defined as the?energy transferred per unit charge
Another measure of energy transfer is work done
Therefore, potential difference can also be defined as the?work done per unit charge
Potential difference is the work done per unit charge
Terminal Potential Differnce & Lost Volts
The?terminal potential difference (p.d)?is the potential difference across the terminals of a cell
If there was no internal resistance, the terminal p.d would be equal to the e.m.f
It is defined as:
V?=?IR
Where:
V?= terminal p.d (V)
I?= current (A)
R?= load resistance (Ω)
If a cell has internal resistance?r, the terminal p.d is always?lower?than the e.m.f
If you have a load resistor?R?across the cell's terminals, then the terminal p.d?V?is equal to?the p.d across the load resistor
In a closed circuit, current flows through a cell and a potential difference develops?across the internal resistance
Since resistance opposes current, this reduces the energy per unit charge (voltage) available to the rest of the external circuit
This difference is called the ‘lost volts’
Lost volts?is usually represented by little?v
It is defined as the voltage?lost?in the cell due to internal resistance
So, from conservation of energy, we can say:
v?= e.m.f ? terminal p.d
v = ε – V =?Ir?(Ohm’s law)
Where:
v?= lost volts (V)
I?= current (A)
r?= internal resistance of the battery (Ω)
ε =?e.m.f (V)
V =?terminal p.d (V)
Therefore, lost volts is the?difference?between the e.m.f and the terminal p.d
Discharging a Cell
When a cell is discharging, it will?not discharge?a?constant?amount?of voltage
Instead, an?initial high amount?that?slowly decreases?over time is discharged?ending?in a?rapid decrease
This means that cells make a?distinctive discharge curve?with a drop, plateau and final rapid drop
Typical discharge curves for a 1.5 V terminal cell showing discharge for a 0.5A, 1A and 1.5A drawing current
The Capacity of a Cell
The?capacity?of a cell is the?amount of charge?that it contains and is?able to discharge
This is measured in Ampere hours (A hr)
When a cell has a certain capacity the amount of?current drawn?from this cell will impact the amount of?time?that it can?run?for
In the image above three different drawing currents are shown for the same 1.5 V cells
The relationship between current drawn and hours of cell lifetime is a simple?linear relationship
As an example: A 100 A.hr capacity battery is able to provide 100 hours of 1A current
However, the same battery when fully charged can give 50 hours of charge for a 2A current or 25 hours for a 4A current
Worked Example
A lamp is connected to a 240 V mains supply and another to a 12 V car battery.Both lamps have the same current, yet 240 V lamp glows more brightly. Explain in terms of energy transfer why the 240 V lamp is brighter than the 12 V lamp.
ANSWER:
Both lamps have the same current, which means charge flows at the same rate in both
The 240 V lamp has 20 times more voltage than the 12 V lamp
Voltage is the energy transferred (work done) per unit charge
This means the energy transferred to each coulomb of charge in the 240 V lamp is 20 times greater than for the 12 V lamp
This makes the 240 V lamp shine much brighter than the 12 V lamp
Exam Tip
Think of potential difference as being the?energy per coulomb?of charge transferred between two points in a circuit
If the exam question states 'a battery of negligible internal resistance', this assumes that e.m.f of the battery is equal to its voltage. Internal resistance calculations will not be needed here.
If the battery in the circuit diagram includes internal resistance, then the e.m.f equations must be used.