= Solution
The positive lamp <power> comes from extra <work> by the primary current source. The secondary <electric current> creates a changing <magnetic flux> through the primary, so maintaining the same prescribed primary <electric current> requires an additional terminal <voltage>. With the preceding winding convention,
$$
v_p=L\frac{dI_p}{dt}+M\frac{dI_s}{dt}.
$$
Using <phasors>, $\widetilde I_s=-i\omega M\widetilde I_p/(R+i\omega L_s)$, hence the primary sees the <complex impedance>
$$
Z_{\rm in}=i\omega L+\frac{\omega^2M^2}{R+i\omega L_s},\qquad
\operatorname{Re}Z_{\rm in}=\frac{\omega^2M^2R}{R^2+(\omega L_s)^2}>0.
$$
It follows directly that
$$
\boxed{\langle P_{\rm source}\rangle=\tfrac12I_0^2\operatorname{Re}Z_{\rm in}=\langle P_R\rangle.}
$$
This is the <reflected impedance> of the secondary load. The <self-inductance> and <mutual inductance> terms exchange stored <magnetic energy>, while the lamp dissipates it. Multiplying the two coupled equations from <Kirchhoff's voltage law> by their respective <electric currents> also gives the instantaneous <conservation of energy> identity
$$
v_pI_p=RI_s^2+\frac{d}{dt}\left(\frac12LI_p^2+MI_pI_s+\frac12L_sI_s^2\right).
$$
The derivative averages to zero in periodic steady state. A fixed primary <current source> supplies the required extra <voltage>; a fixed <voltage source> would generally change its primary <electric current> when the lamp is connected.
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