solved by m. bilal azam
TRANSCRIPT
Solved by M. Bilal Azam
Solved by M. Bilal Azam
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Solved by M. Bilal Azam
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Solved by M. Bilal Azam
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
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Bilal Sohaib
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Bilal Sohaib
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Bilal Sohaib
Note
80= 10logP 30 =10 logP´ Subtracting 80-30 = 10 logP/P´ so 5 = log P/P´ P/P´= 10^5
Solved by M. Bilal Azam
Bilal Sohaib
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Bilal Sohaib
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Bilal Sohaib
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Bilal Sohaib
Note
In radians fr=1/√LC
Bilal Sohaib
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Bilal Sohaib
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Bilal Sohaib
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Bilal Sohaib
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Bilal Sohaib
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Bilal Sohaib
Note
total cost = power (kW) × time (h) × cost per unit. total cost = 2 × 2.5 × 30 × 5 =750 Rs
Bilal Sohaib
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Bilal Sohaib
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Bilal Sohaib
Note
Since each resistance is of 12 ohms and I1 is the current passing through R_A and R_B, so I1=V/R=12/24=0.5 A And I2 is the current passing through R_C I2=V/R=12/12 = 1 A Thus power through R_A and R_B is P1=(I^2)R=(0.5)^2*24= 6 W i.e. 3 W in each R_A and R_B. And P2=(I^2)R=(1)^2*12 =12 W in R_C.
Bilal Sohaib
Typewriter
In radians
Bilal Sohaib
Typewriter
In degrees
Solved by M. Bilal Azam
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
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Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Solved by M. Bilal Azam
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Solved by M. Bilal Azam
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle
Bilal Sohaib
Rectangle