The secondary toils of many transformers have tapr, these arc tra Connections to an intermediate part of the secondary1 coil, so that, in addition to the full secondary voltage, a fixed fraction of the secondary voltage is also available. Consider a transformer with a 115 - V AC primary and a 12 - V AC secondary that has a center tap. The 12 - V AC secondary voltage is applied to a load that dissipates an average power of 4.0 W, and the center tap (6.0 V AC) to a load of average power 12 W. What is the current in the primary?
Explanation / Answer
Total power delivered tosecondarycoil PS = 12+4.0 = 16.0 W For an idelatransformer PS = Powersupplied to primarycoil = 16.0 W VP *IP = 16.0 Current through primarycoil IP = 16.0/115 = 0.139 A