JEE Mains · Physics · STD 12 -7. Alternating current
In a scries \(LR\) circuit, power of \(400\, W\) is dissipated from a source of \(250\, V , 50\, Hz\). The power factor of the circuit is \(0.8 .\) In order to bring the power factor to unity, a capacitor of value \(C\) is added in series to the \(L\) and \(R\). Taking the value of \(C\) as \(\left(\frac{ n }{3 \pi}\right) \mu F ,\) then value of \(n\) is\(......\)
- A \(200\)
- B \(250\)
- C \(350\)
- D \(400\)
Answer & Solution
Correct Answer
(D) \(400\)
Step-by-step Solution
Detailed explanation
\(P =\frac{ E _{ rms }^{2}}{ Z } \cos \phi\) \(400=\frac{(250)^{2} \times 0.8}{ Z }\) \(Z =25 \times 5=125\) \(X _{ L }=125 \sin \phi=125 \times 0.6=75\)
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