In inductance of `(4//pi)` H and the resistor R, are connected in series and an alternating emf of frequency 50 Hz is applied across combination. If p

In inductance of `(4//pi)` H and the resistor R, are connected in series and an alternating emf of frequency 50 Hz is applied across combination. If phase difference between applied emf and current is `45^(@)` then the value of R is
A. `200Omega`
B. `400Omega`
C. `600Omega`
D. `800Omega`

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

Salim Ahmed Kawsar

Correct Answer - B
`tanphi=(X_(L))/(R)`
`therefore R=(2pifL)/(tanphi)`
`=(2pixx50xx4)/(tan45xxpi)=400Omega`

6 views Answered 2023-02-05 15:29

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