Please answer with good writing and not briefly You have a battery. In order to

Question

Please answer with good writing and not briefly

You have a battery. In order to determine its properties, you connect some resistors to it and measure the current that pass through. (See the picture in the 1st question.) If you connect a R = 2 ohm resistor, the current is measured as 0.50 A. If the resistor is R = 4 ohm, the current becomes 0.30 A. (a) Find the emf and the internal resistance of the battery. (b) If you connect a R = 1 k ohm resistor to the battery, the potential drop on the resistor, V_R, will still be smaller than the emf, element, of the battery, albeit by a very small amount. Find the fractional difference between V_R and element.

Explanation / Answer

current through resistor I = E/(R + r)

for R1 = 2 ohm

I1 = E/(R1 + r)

for R2 = 4 ohms

I2 = E/(R2 + r)

I1 = 0.5 A

I2 = 0.3 A

I2/I1 = (R1 + r)/(R2 + r)

0.3/0.5 = (2 + r)/(4 + r)

internal resistance r = 1 ohm

E = I1*(R1 + r )

emf E = 0.5*(2+1) = 1.5 V

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(b)

current I = E/(R + r)

I = 1.5/(1000+1)

I = 0.001498 A

VR = I*R = 0.001498*1000 = 1.498 V

fractioanl difference = 1.5 - 1.498 = 0.002 V

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