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27. Jan 2023•0 gefällt mir•2 views

27. Jan 2023•0 gefällt mir•2 views

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8. What is the Maximum power that can be delivered to the lamp in the circuit below? What will the turns ratio be? How much power is lost to the source resistance? Solution Here in the circuit th turns ratio is given as 0.7. The maximum current flowing through the primary of the transformer is (28/10)=2.8a. Now from the standard transformer equation we have , N2/N1=I1/I2= so we have i2= 2.8/0.7=4A. so the maximum power that can be delivered to the load with this rating is I 2 R= 4 2 *4.9= 78.4WATTS. SO the power lost due to resistance source= IR=(2.8*10)=28v. .

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- 1. 8. What is the Maximum power that can be delivered to the lamp in the circuit below? What will the turns ratio be? How much power is lost to the source resistance? Solution Here in the circuit th turns ratio is given as 0.7. The maximum current flowing through the primary of the transformer is (28/10)=2.8a. Now from the standard transformer equation we have , N2/N1=I1/I2= so we have i2= 2.8/0.7=4A. so the maximum power that can be delivered to the load with this rating is I 2 R= 4 2 *4.9= 78.4WATTS. SO the power lost due to resistance source= IR=(2.8*10)=28v.