I=ℇ/(R+r)
R= (ℇ-Ir)/I= (7.9-0.88•1.37)/0.88 = …
P=I²r = …
A battery with an emf of 7.9 V and internal resistance of 1.37 Ω is connected across a load resistor
R.If the current in the circuit is 0.88 A, what is the value of R?
Answer in units of Ω
What power is dissipated in the internal resistance of the battery?
Answer in units of W
How do you set this problem up?
2 answers
i = 7.9/(1.37+R)
.88 = 7.9/(1.37+R)
1.37 + R = 8.98
so
R = 7.61 ohms
P = i^2 r = .88^2(1.37) = 1.06 Watts
.88 = 7.9/(1.37+R)
1.37 + R = 8.98
so
R = 7.61 ohms
P = i^2 r = .88^2(1.37) = 1.06 Watts