# 6.08 Example questions involving Internal resistance.

Example 1.
A battery is connected to a 98 W load resistor, causing 60mA (0.06A) to flow . When the battery is disconnected from the circuit, its terminal voltage is measured to be 6V.
Determine the e.m.f. of the battery, its internal resistance and the terminal voltage, when the 98 W resistor is connected. • the e.m.f. voltage is equal to the open circuit terminal voltage of the battery = 6V .
• r = E/I - RL .
• r = 6/(60 x 10-3) - 98 = 2W .
• V = E - Ir .
• V = 6 - 60 x 10-3 x 2 = 5.88V .

Example 2.
A battery has an e.m.f. of 10V and an internal resistance of 0.5W, calculate the terminal voltage when the battery is connected to:
a) A 10W resistor.
b) A 50 W resistor.
c) A 100W resistor. a)

• RT = r + RL .
• 0.5 + 10 = 10.5W .
• I = E/RT .
• 10/10.5 = 0.952A .
• V = E - Ir .
• 10 - 0.952 x 0.5 = 9.52 V .

b)

• RT = r + RL .
• 0.5 + 50 = 50.5W .
• I = E/RT .
• 10/50.5 = 0.198A .
• V = E - Ir .
• 10 - 0.198 x 0.5 = 9.9 V .

c)

• RT = r + RL .
• 0.5 + 100 = 100.5W
• I = E/RT .
• 10/100.5 = 0.0995A
• V = E - Ir .
• 10 - 0.0995 x 0.5 = 9.95 V .