This question involves the concepts of turns ratio and power.
a. The voltage in the secondary circuit is "2.17 V".
b. The power drawn by the primary circuit is "2.71 watt".
c. Power supplied to the secondary circuit is "2.71 watt".
From the turns ratio formula we know that:
[tex]\frac{N_p}{N_s}=\frac{V_p}{V_s}\\\\V_s = \frac{N_sV_p}{N_p}[/tex]
where,
Therefore,
[tex]V_s=\frac{(20\ turns)(325\ V)}{3000\ turns}[/tex]
[tex]V_s = 2.17\ V[/tex]
Electrical power is given by the following formula:
[tex]P_p = I_pV_p[/tex]
where,
Therefore,
[tex]P_p=(1.25\ A)(2.17\ V)\\\\P_p = 2.71\ watt[/tex]
Since the transformer is perfectly efficient. It means that the total power drawn by the primary circuit is supplied to the secondary circuit.
[tex]P_s=P_p\\\\P_s=2.71\ watt[/tex]
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