Answer:
(a) 5.725 Ω
(b) 1.3 Ω
Explanation:
(a)
E = emf of the battery = 15.0 Volts
V = terminal voltage of the battery = 12.2 Volts
P = Power delivered to external load resistor "R" = 26.0 W
R = resistance of external load resistor
Power delivered to external load resistor is given as
[tex]P = \frac{V^{2}}{R}[/tex]
26.0 = 12.2²/R
R = 5.725 Ω
(b)
r = internal resistance of the battery
i = current coming from the battery
Power delivered to external load resistor is given as
P = i V
26.0 = i (12.2)
i = 2.13 A
Terminal voltage is given as
V = E - ir
12.2 = 15 - (2.13) r
r = 1.3 Ω
