A battery with an emf of 3.00 V has an internal resistance r. When connected to

Question

A battery with an emf of 3.00 V has an internal resistance r. When connected to a resistor R, the terminal voltage is 2.90 V and the current Is 0.20 A. What is the value of the external resistor R? Answer Tries 0/8 What is the internal resistance r of the battery? Tries 0/8 What is the energy dissipated in the battery's internal resistance in 2.4 minutes? Tries 0/8 When a second identical battery is added in series and the external resistor is R = 18 Ohms what is the resulting current? Tries 0/8

Explanation / Answer

given dataemf =3V
internal resistance =r
terminal voltage =2.9 V
I=0.20A
R=?
r=?
energy dissipated in t=2.4 minutes
if R=18 ohms then I=?

B) internal resistance r=(emf-terminal voltage )/ current =( 3-2.9)/0.20 =0.5 ohm

A)external resisitance R

emf / current = R+r

3/0.20 = R + 0.5

R=14.5 ohms

C)

energy dissipated

E=I^2 r*t

= 0.20*0.20*0.5*2.4*60 = 2.88 J

D)Resulting current

2V /(18+2*r) = 2*3 / (18+2*0.5) = 0.315 A

  

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