1}The power factor of a series RL circuit is 1/√2.if the frequency if the applied ac is doubled,the power factor becomes
1)1/√3
2)1/√5
3)1/√7
4)1/√11
2}A circuit has a resistance of emf 200v is appplief to a circuit consisting if a resistance of 40ohm.they r connected in series with an ac source.if the peak value of current is 1A nd the voltage is 220v,the power consumed by the circuit is
1)66w
2)6.6w
3)0.66w
4)33w
kvnmurty:
Qn 2... What is 200V?? Can you write correct problem here again..
Correct statement of qn 2.please
The ques i wrote as it was given sir
Answers
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X_L = ω L
Power factor of a RL A.C. Circuit is given by
p.f. = R /√[ R² + X_L²]
= 1 /√[ 1 + (ω²L²/R²) ] = 1/√2 given
=> ωL/R = 1
If frequency is doubled, then ωL/R becomes twice.
p.f. = 1/√[1 + 4] = 1/√5
==========
2 ) This problem statement is not clear...
Power factor of a RL A.C. Circuit is given by
p.f. = R /√[ R² + X_L²]
= 1 /√[ 1 + (ω²L²/R²) ] = 1/√2 given
=> ωL/R = 1
If frequency is doubled, then ωL/R becomes twice.
p.f. = 1/√[1 + 4] = 1/√5
==========
2 ) This problem statement is not clear...
clik on red heart thanks above pls
2nd 1 we can solve
How do u solve it if problem statement is not clear
Some data is missing
No sir so i too was confused abt it
how do you accept the line in the answer that R = 30 ohms ? how could you solve qn 2 ? you say you can solve it
thank you @deeksha for selecting brainliest. that is very nice of you.
U explained it so perfectly sir..
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