An ideal transformer, connected to an electrical device, is plugged into a 220 V source. Ifthe internal resistance of the device is 28.0 Ω and a current of 495 mA is going throughit, how much current is drawn from the source?
4 years ago
Answered By Alexandra G
Transformers obey the law of conservation of energy, so the energy going into the primary coil must equal the energy coming out of the secondary coil. We know that Energy = Power x Time and Power = Voltage x Current. Combining the law of conservation of energy and these equations, we know that:
Vprimary*Iprimary = Vsecondary*Isecondary
From your question (assuming I am reading and understanding it correctly), the source would be the primary coil and we are given the primary voltage as 220 V (Vprimary = 220 V). We need to find the current drawn from the source, i.e. need to find Iprimary.
On the secondary side, I assume this is the electrical device that the transformer is connected to. We are given the current as 495 mA (so Isecondary = 0.495 A) and the resistance as 28 Ohms. Using Ohm's Law (V=IR), we can solve for the secondary voltage as Vsecondary = Isecondary*Rsecondary.
Once we have solved for the secondary voltage, you can use this in the conservation of energy equation and solve for the primary current, which is the current drawn from the source.
Hope this helps! Please reach out with any other questions or if I have misunderstood something.
4 years ago
Answered By Alexandra G
Transformers obey the law of conservation of energy, so the energy going into the primary coil must equal the energy coming out of the secondary coil. We know that Energy = Power x Time and Power = Voltage x Current. Combining the law of conservation of energy and these equations, we know that:
Vprimary*Iprimary = Vsecondary*Isecondary
From your question (assuming I am reading and understanding it correctly), the source would be the primary coil and we are given the primary voltage as 220 V (Vprimary = 220 V). We need to find the current drawn from the source, i.e. need to find Iprimary.
On the secondary side, I assume this is the electrical device that the transformer is connected to. We are given the current as 495 mA (so Isecondary = 0.495 A) and the resistance as 28 Ohms. Using Ohm's Law (V=IR), we can solve for the secondary voltage as Vsecondary = Isecondary*Rsecondary.
Once we have solved for the secondary voltage, you can use this in the conservation of energy equation and solve for the primary current, which is the current drawn from the source.
Hope this helps! Please reach out with any other questions or if I have misunderstood something.