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Answer:
a) 0.1832 A
b) 11.91 Volts
c) 2.18 Watt , 0.0168 Watt
Explanation:
(a)
R = external resistor connected to the terminals of the battery = 65 Ω
E = Emf of the battery = 12.0 Volts
r = internal resistance of the battery = 0.5 Ω
i = current flowing in the circuit
Using ohm's law
E = i (R + r)
12 = i (65 + 0.5)
i = 0.1832 A
(b)
Terminal voltage is given as
= i R
= (0.1832) (65)
= 11.91 Volts
(c)
Power dissipated in the resister R is given as
= i²R
= (0.1832)²(65)
= 2.18 Watt
Power dissipated in the internal resistance is given as
= i²r
= (0.1832)²(0.5)
= 0.0168 Watt