Calculating the epislon ε and current intensity? help please

[phy_freak]

Homework Statement

a battery connected to a light bulb that has electrical resistance of (6 Ω) and the current intensity that gets through the bulb counts (12 A), if the internal resistance of the battery was (2 Ω):-
1- what is the Driving force of the battery (ε)?
2- if the light bulb was exchanged with another bulb that has resistance of (4 Ω) what would the current intensity getting through equal?

Homework Equations

ε= VR + Vr
ε=I(R+r)
ε=W/q
r=ε-VR / I
VR = I×R
Vr = I×r
(there might be other forms of the same equations)


I actually don't know weather the electrical resistance given in the problem was r, R, VR or Vr

 

[tiny-tim]

Hi phy_freak!

I actually don't know weather the electrical resistance given in the problem was r, R, VR or Vr

In this case, it's the inital letter that tells you …

r or R is for Resistance,

v or V is for Voltage (= electric potential = emf = driving force)

(and i suppose I is for Intensity of current, and W is for energy because it stands for Work, which is a form of energy)

 

[phy_freak]

Hi phy_freak! In this case, it's the inital letter that tells you …

r or R is for Resistance,

v or V is for Voltage (= electric potential = emf = driving force)

(and i suppose I is for Intensity of current, and W is for energy because it stands for Work, which is a form of energy)

so we use the equation: ε= VR + Vr to solve the first question? then it means that VR=6 (the given bulb's resistance) and Vr=2 (battery given resistance)
so ε=6+2 =8V? is that correct?

 

[tiny-tim]

so we use the equation: ε= VR + Vr to solve the first question?

No, you're misunderstanding what this equation is.

It's ε= V_R + V_r,

which in words means the emf (driving force) equals the voltage drop (electric potential difference) across the resistance R plus the voltage drop across the resistance r.

R and r in this equation aren't factors, to be multiplied … they're only labels, to tell you which V you're talking about.

(so if the resistances were called R₁ and R₂, the voltage drops would be called V₁ and V₂, with the labels "1" and "2")

The equation you need is ε = I(R+r) …

(because ε= V_R + V_r, and V_R = IR and V_r = Ir, so altogether ε = IR + Ir = I(R+r))

 

[phy_freak]

i used ε = I(R+r), assuming that R=6 (the bulb's resistance) and r=2 (battery resistance), ε equaled 96 V.

and i used I=ε/R+r (with the ε equaling 96) for the second question, (I) equaled 16A

are those correct?

 

[tiny-tim]

i used ε = I(R+r), assuming that R=6 (the bulb's resistance) and r=2 (battery resistance), ε equaled 96 V.

and i used I=ε/R+r (with the ε equaling 96) for the second question, (I) equaled 16A

are those correct?

Yup!

 

[phy_freak]

thank you very much you were helpful :)