A car battery has a terminal potential difference of 12.712.7 V when there is no load. When the starter motor draws 8080 A, the terminal potential difference drops to 1010 V. What is the internal resistance of the battery?
_____ Ω
Physic's Question, Please answer?
2016-03-02 1:23 am
回答 (2)
2016-03-02 1:34 am
e.m.f. = 12.7 V
When I = 80 A, V = 10 V
r = ? Ω
V = I R
10 = 80 × R
R = 0.125 Ω
e.m.f. = I × (R + r)
12.7 = 80 × (0.125 + r)
10 + 80r = 12.7
80r = 2.7
r = 0.034 Ω
When I = 80 A, V = 10 V
r = ? Ω
V = I R
10 = 80 × R
R = 0.125 Ω
e.m.f. = I × (R + r)
12.7 = 80 × (0.125 + r)
10 + 80r = 12.7
80r = 2.7
r = 0.034 Ω
2016-03-02 1:31 am
A battery is modeled as a voltage source (value = no load terminal voltage) in series with an "internal resistance". In this case the voltage source behind the internal resistance has value 12.7V. Let the internal resistance in series with this 12.7V source = Ri. The applicable equation is:
12.7v - 80A*Ri = 10v -----> (12.7 - 10)/80 = 2.7/80 = Ri = 0.03375Ω = 27/800 Ω
Ri = 0.03375Ω which would be 0.034Ω with 2 significant figures.
12.7v - 80A*Ri = 10v -----> (12.7 - 10)/80 = 2.7/80 = Ri = 0.03375Ω = 27/800 Ω
Ri = 0.03375Ω which would be 0.034Ω with 2 significant figures.
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