Internal Resistance (Cambridge International AS Physics)

Spec point — 10.1.5

Internal resistance

• All power supplies have some resistance between their terminals

  • This is called internal resistance (r)

• This internal resistance causes the charge circulating to dissipate some energy from the power supply itself

  • This is why the cell becomes warm after a period of time

• The internal resistance therefore causes a loss of voltage or energy loss in a power supply

• A cell can be thought of as a source of e.m.f with an internal resistance connected in series. This is shown in the circuit diagram below:

Circuit with e.m.f

Circuit showing the e.m.f and internal resistance of a power supply

• V_R is the terminal potential difference

  • This is the voltage available in the circuit itself
  • Terminal p.d = I × R(Ohm’s law)

• When a load resistor is connected, current flows through the cell and a potential difference develops across the internal resistance. This voltage is not available to the rest of the circuit so is called the ‘lost volts’

• V_r is the lost volts

  • This is the voltage lost in the cell due to internal resistance, so, from conservation of energy:
  • Lost volts = e.m.f − terminal p.d
  • Lost volts = I × r(Ohm’s law)

• The e.m.f is the sum of these potential differences, giving the equation below:

$E = I R + I r = I ( R + r )$

• Where:

  • E = e.m.f (V)
  • I = current (A)
  • R = load resistance (Ω)
  • r = internal resistance (Ω)

• IR is collectively known as the 'terminal potential difference'

• *Ir *is collectively known as the 'lost volts'

Worked Example A battery of e.m.f 7.3 V and internal resistance r of 0.3 Ω is connected in series with a resistor of resistance 9.5 Ω.

Determine:

a) The current in the circuit

b) Lost volts from the battery

Answer:

a)

Step 1: List the known quantities:

• E.m.f, E = 7.3 V
• Load resistance, R = 9.5 Ω
• Internal resistance, r = 0.3 Ω

Step 2: Use the e.m.f equation to determine the current I

$E = I ( R + r )$

$I=\frac{E}{R+r}$

Step 3: Substitute the values

$I=\frac{7.3}{9.5+0.3}=0.7449A\approx0.74A$

b)

Step 1: State the equation for lost volts

• The lost volts are the voltage lost due to internal resistance

$\text{lost volts}=Ir$

Step 2: Substitute the values

$0.7449\times0.3=0.22V$