Suppose we measure the terminal voltage of a rechargeable NiCad battery and find

Question

Suppose we measure the terminal voltage of a rechargeable NiCad battery and find it to be 1.28 V when measured by a voltmeter. (Remember: voltmeters have very high resistance.) We now connect a 8.0-ohm resistor across the terminals, and find that the terminal voltage has now changed, to 1.16 V. Find the internal resistance of the battery.

I am trying to use 1.28V - 1.16 V = 0.12V which is the votage across to the 8.0-ohm resistor.

then get the current by 0.12V / 8 = 0.015A

1.16 V / 0.015 A = 77.33 ohm

but the answer is incorrect, can someone help?

Explanation / Answer

Le the internal resistance be r

connected resistor = R = 8 ohms

Le the source voltage is E

1.28 = E

1.16 = E - i2 r = i2 R

i2 = 1.16/8 = 0.145 A

1.16 = 1.28 - 0.145 * r

r = ( 1.28 - 1.16) / 0.145 = 0.827 ohms

internal resistance of the battery = 0.827 ohms

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