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This is chapter 27 problem number 48.
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We have a dc motor with internal resistance of 3 .3 .2 oms.
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This is internal resistance.
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And the potential supplied to the motor supplied to the motor is 120 volts.
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However, in the emf, in the rotor is 105 volts due to this internal resistance.
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So in part a, we're trying to calculate the current drawn by the motor.
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So we can have the current here in part a.
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Well, in order to get there, of course, the supplied voltage has to be equal to the emf that we have in the rotor, plus the potential across this resistor, right? internal resistance.
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So then here, if you're trying to calculate the current, then the sub -wide minus the emf divided by the internal resistance is going to give us the current drawn by the motor.
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So then 120 minus 105 divided by 3 .2.
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It's going to give us 4 .69 amps...