Syllabus Edition

First teaching 2014

Last exams 2024

|

Terminal Potential Difference (DP IB Physics: SL)

Revision Note

Katie M

Author

Katie M

Last updated

Terminal Potential Difference

Defining Potential Difference

  • 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

Potential difference in a circuit, downloadable AS & A Level Physics revision notes

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 equation, downloadable AS & A Level Physics revision notes

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

5-3-3-discharge-curve_sl-physics-rn

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.WE - Potential difference question image, downloadable AS & A Level Physics revision notes

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

Examiner 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.

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.