Answer
$0.36A,0.16KV$
Work Step by Step
We can find the primary current and its voltage as follows:
$I_p=(I_s)(\frac{N_p}{N_s})$
We plug in the known values to obtain:
$I_p=(12mA)(\frac{750}{25})=360mA=0.36A$
Now the primary voltage is given as
$V_p=(V_s)(\frac{N_p}{N_s})$
We plug in the known values to obtain:
$V_p=(4800)(\frac{25}{750})$
$V_p=160V=0.16KV$
