A series LCR circuit is connected across a source E of e.m.f. E=15...

Given the equations for e.m.f. and current:

$E = 15\sin \left(50\pit - \dfrac{\pi}{3}\right)<br><br>$ I = 5\sin \l \left(50\pi$t +$\dfrac{\pi}{6}\right)

The peak voltage is $E_0 = 15$ V and the peak current is $I_0 = 5$ A.

The impedance of the circuit is given by:

$Z =$\dfrac{E_0}{I_0} = \dfrac{15}{5} =3\ \Omega$<br><br>The phase of the e.m.f. is$\phi_E = 50\pi$t -$\dfrac{\pi}{3}$and the phase of the current is\phi_I =$50\pi$t +$\dfrac{\pi}{6}.

The phase difference \Delta \phi between$the$e.m.f. and the current is:

\Delta \phi = |\phi_I - \phi_E| = \left|\lleft(50\pit + \dfrac{\pi}{6}\right) - \lleft(50\pi$t -$\dfrac{\pi}{3})|

\Delta \phi = \dfrac{\pi}{6} + \dfrac{\pi}{3} = \dfrac{\pi}{6} + \dfrac{2\pi}{6} = \dfrac{3\pi}{6} = \dfrac{\pi}{2} radians.

Thus,$the$impedance is 3\ \Omega$and the phase difference is \left(\dfrac{\pi}{2}\right)^{c}$.

Answer: $3\ \Omega,\ \left(\dfrac{\pi}{2}\right)^{c}$