Question

The voltage across the terminals of a 9.0 v battery is 8.5 v when the battery is connected to a 60 ω load. part a what is the battery's internal resistance?

Answer 1

V = terminal Voltage = potential difference across the load = 8.5 Volts

E = EMF of the battery = 9.0 Volts

P = Power rating of the load = 60 Watt

i = current flowing through the circuit = current flowing through the load

Power through the load is given as

P = i V

inserting the values

60 = i (8.5)

i = 7.06 A

Terminal voltage is given as

V = E - i r                                      where r = internal resistance

8.5 = 9 - (7.06) r

r = 0.071 ohm

Answer 2

Refer to the diagram shown below.

i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω

Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5         (10

Also,
R₂*i = 9.5         (2)

Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A

From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)

Answer: 0.08 Ω