Use the exact values you enter in previous answer(s) to make later calculations.

Question

Use the exact values you enter in previous answer(s) to make later calculations.

The battery in the figure below is a real battery. That is, it has some internal resistance Rint in series with an ideal battery with an emf of V0. In an attempt to determine Rint and V0, a resistor

R1 = 705

is first placed across points A and B, and the voltage across the resistor is measured to be 5.5 V. This resistor is then replaced with a second resistor with

R2 = 2,300 ,

and the voltage across R2 is found to be 9.8 V. Find Rint and V0.

Explanation / Answer

First case

R1 = 705

VR= 5.5V

Here  internal resistance Rint in series with an ideal battery with an emf of V0

Here for series one I remains same

So we get I1= VR/ R1=5.5V/705=0.0078A

V0=I1[R1+Rint]

Second case

R1 = 2300

VR= 9.8V

Here for series one I remains same

So we get I2= VR/ R1=9.8V/2300=0.0043A

EMF=I(R+Rint) , Rint is internal resistance, R is variable resistor

EMF 1 = EMF 2

I1(R1 + Rint) = I 2 (R2 +Rint )

0.0078(705+ Rint) = 0.0043(2300+Rint )

5.499+0.0078 Rint=9.89+0.0043Rint

0.0035Rint=4.391

Rint= 1254.6

V0=I1[R1+Rint]=0.0078[705+1254.6]=15.28V

V0=15.28V

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