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Internal resistance?
What potential difference is measured across a 14.1 Ω load resistor when it is connected across a battery of emf 3.48 V and internal resistance 0.486 Ω?
3 Answers
- billrussell42Lv 71 month agoFavorite Answer
total R = 14.1 + 0.486 = 14.586
I = E/R = 3.48/14.586 = 0.2386 amps
Voltage across load = IR = 0.2386 x 14.1 = 3.36 volts
- oldschoolLv 71 month ago
The model for a battery is an Emf in series with an internal resistance Ri. This is illustrated by switching your headlights on with engine off. The lights dim when you start the engine. They brighten after starting thanks to the alternator boosting the voltage. The equation for the model ckt is
Emf - Iout*Ri = Vout
Where Iout is the current out of the battery into the load and Ri is the internal resistance of the battery.
3.48 - Iout*0.486 = Vout = Iout*14.1
3.48 = Iout*(0.486+14.1)
Iout = 3.48/14.586 = 0.239 = 239mA
Vout = 0.239 * 14.1 = 3.36V <<<<
Hope this helps.
