JEE Mains · Physics · STD 12 -7. Alternating current
An inductance coil has a reactance of \(100\, \Omega\). When an \(AC\) signal of frequency \(1000\, Hz\) is applied to the coil, the applied voltage leads the current by \(45^{\circ}\). The self- inductance of the coil is
- A \(1.1 \times 10^{-2}\; H\)
- B \(1.1 \times 10^{-1} \;H\)
- C \(5.5 \times 10^{-5} \;H\)
- D \(6.7 \times 10^{-7}\; H\)
Answer & Solution
Correct Answer
(A) \(1.1 \times 10^{-2}\; H\)
Step-by-step Solution
Detailed explanation
Reactance of inductance coil \(=\sqrt{{ R ^{2}+ x _{ L }^{2}}}=100\) \(...(i)\) \(f =1000 Hz\) of applied \(AC\) signal Voltage leads current bly \(45^{\circ}\) \(\tan 45^{\circ}=\frac{ i X _{ L }}{ iR }=\frac{\omega L }{ R }\) ie \(R = X _{ L }=\omega L\) Putting in eqn…
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