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The no-load output voltage of a DC power supply is measured at 15V When a 600 ohms load is connected to the output the output drops to 13.7V Calculate the internal resistance of the power supply?
Wiki User
∙ 12y ago
This question can be answered using voltage dividers. Assume the power supply consists of a voltage source and a resistor. With no load, all of the voltage source's voltage is dissipated by the internal resistor of 15V. When there is a load, there are two resistors in series. To calculate the internal resistance:
1. I=V/R. You know the 600ohm resistor dissipated 13.7V. So that would mean a current of 13.7/600=22.8mA
2. If the 600ohm resistor dropped 13.7, kirchoff's voltage law would tell us the internal resistor dropped 15-13.7=1.3V.
3. R=V/I, Use the current to calculate the internal resistance. 1.3/22.8mA = 56.9ohms
CommentFurther to the above answer, a voltage-source's voltage is not 'dissipated by the internal resistance when on no load'. On no load, there is no current passing through the internal resistance, so no 'voltage dissipation' can takes plac -i.e. the non-load voltage is 15 V.
Wiki User
∙ 12y ago
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what is the diference betwean calculated and maesured value
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What happen when ammeter connected in parallel?
an ideal ammeter has zero or negligible resistance when this is connected in series no effective resistance would be added in the circuit so that the value of curret that we get is exactly of the circuit only. but when the ammeter is connected in parllel as it has zero resistance , the resistor to which it is connected in parllel gets shorted and due to his the effective resistance of the circuit is changed and so the effective current ... due to this the w=value measured by the ammeter would be different (incresed due to dec. in effective resistance)
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it is measured in Ohmmeter Ohms
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What is a fixed resistor measured in?
Resistance is measured in ohms.
What is resistance expressed in?
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