Question

A series LCR circuit is connected across a source E of e.m.f. \(E=15\,\sin \left(50\pi t-\frac{\pi }{3}\right)\). The current from the supply is \(I=5\,\sin \left(50\pi t+\frac{\pi }{6}\right)\). The impedance of the circuit and the phase difference between e.m.f. and current are, respectively, [all the quantities in the above two equations are in SI units.]

• A. \(5\ \Omega ,\ \left(\frac{\pi }{6}\right)^{c}\)
• B. \(3\ \Omega ,\ \left(\frac{\pi }{2}\right)^{c}\)
• C. \(7\ \Omega ,\ \left(\frac{\pi }{12}\right)^{c}\)
• D. \(9\ \Omega ,\ \left(\frac{\pi }{4}\right)^{c}\)

Hint:

Impedance is the ratio of peak voltage to peak current. Phase difference is the difference between the two phase angles.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Impedance \(Z=\dfrac{E_{0}}{I_{0}}=\dfrac{15}{5}=3\ \Omega\).

Step 2: Phase difference
Phase of emf \(=-\dfrac{\pi}{3}\) and phase of current \(=+\dfrac{\pi}{6}\).
\(\Delta\phi=\dfrac{\pi}{6}-\left(-\dfrac{\pi}{3}\right)=\dfrac{\pi}{2}\)

Step 3: Result
The current leads the emf by \(\dfrac{\pi}{2}\) rad.

Final Answer:
The impedance is 3 ohm and the phase difference is \(\frac{\pi}{2}\) rad. \[ \boxed{\text{(B) }3\ \Omega,\ \left(\dfrac{\pi}{2}\right)^{c}} \]