Question

A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, Vab, and (c) the power dissipated in the resistor R and in the battery’s internal resistance r.

Answer 1

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

$V_{ab}$ = i R

$V_{ab}$ = (0.1832) (65)

$V_{ab}$ = 11.91 Volts

(c)

Power dissipated in the resister R is given as

$P_{R}$ = i²R

$P_{R}$ = (0.1832)²(65)

$P_{R}$ = 2.18 Watt

Power dissipated in the internal resistance is given as

$P_{r}$ = i²r

$P_{r}$ = (0.1832)²(0.5)

$P_{r}$ = 0.0168 Watt