解$:(1)I_{额}=\frac {P_{额}}{U_{额}}=\frac {9\ \mathrm {W}}{220\ \mathrm {V}}≈0.04\ \mathrm {A}$
$(2)W_{LED}=P_{LED}t=9×10^{-3}\ \mathrm {kW}×10\ \mathrm {h}=0.09\ \mathrm {kW·h}$
$W_{白炽灯}=P_{白炽灯}t=60×10^{-3}\ \mathrm {kW}×10\ \mathrm {h}=0.6\ \mathrm {kW·h}$
$W_{差}=W_{白炽灯}-W_{LED}=0.6\ \mathrm {kW·h}-0.09\ \mathrm {kW·h}=0.51\ \mathrm {kW·h}$