B-005
B-005
Section B-005-011
A transformer with a 120-volt primary voltage supplies 250 watts to a transmitter. Neglecting losses, what is the approximate primary current?
The power needed by the primary is the power delivered by the secondary plus the losses in the transformer. The 250 watts must be drawn from the source where the primary is connected. The Power Law: P = E * I, power is voltage times current. Transformed to solve for I, it becomes I = P / E. In this example, I = 250 watts / 120 volts.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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How can a transformer with two windings change impedance?
Impedance (Z) can be seen as the ratio of voltage to current in AC circuits: Z = E / I. While functioning with different ratios of voltage to current, the primary and secondary windings of a transformer can operate at different impedances.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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A transformer with a single 12-volt secondary draws 0.5 amperes through its 120-volt primary. Assuming no losses, what current is drawn from the secondary?
As work is performed at a lower voltage on the secondary side, current on the secondary is larger. The turns ratio is '10 to 1' ( 120 volts to 12 volts ), the current ratio follows the inverse of that ratio: 10 * 0.5 amperes = 5 amperes. Method B: Primary consumes 60 watts ( 120 volts * 0.5 amperes ), secondary must draw that same power (discounting losses). What is the secondary current for 60 watts at 12 volts ? I = P / E (derived from P = E * I), I = 60 watts / 12 volts = 5 amperes.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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The primary winding of a transformer has 250 turns, and the secondary has 500 turns. If the input voltage is 120 volts, what is the secondary voltage?
A 'step-up' transformer, the secondary uses twice as many turns as the primary, voltage is doubled ( exactly per the turns ratio ).
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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The strength of the magnetic field around a conductor in air is:
Current and magnetism are closely related: current in a conductor sets up a magnetic field, dropping a conductor through magnetic lines of force creates a current. The voltage would only be of concern for an electrical field. Reference to the conductor's diameter is a useless clue.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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Maximum induced voltage in a coil occurs when:
For induction to take place in a wire, a conductor must be subjected to a moving magnetic field (no movement, no induction). Either the conductor must move in the magnetic field OR the magnetic field must move if the conductor is immobile. If current changes drastically within a short period of time ('rate of change'), the magnetic field around the conductor changes rapidly, induction is maximized.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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A transformer primary winding consumes 10 watts. Neglecting losses, if the secondary voltage is 5 volts, what is the secondary current?
Excluding losses, the power consumed by the primary equals the power required of the secondary. The Power Law: P = E * I, power is voltage times current. Transformed to solve for I, it becomes I = P / E. In this example, I = 10 watts / 5 volts.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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A step-up transformer with a primary to secondary turns ratio of 1:5 delivers 50 milliamperes to a load. Assuming 100% efficiency, what is the primary current?
A turns ratio of '1 to 5' indicates a 'step-up' transformer, primary current will be larger than the secondary current by the inverse of that ratio. In this example, primary current is 5 * times 50 mA = 250 milliamperes = 0.25 amperes. Transformers do not "create" power out of nothing, the power ( E * I ) flowing into the primary equals the power drawn by the secondary plus losses (which are ignored for the sake of simplicity). For power to remain "comparable" on both sides of the transformer, current goes up if voltage increases and vice versa.
Original copyright; explanations transcribed with permission from Francois VE2AAY, author of the ExHAMiner exam simulator. Do not copy without his permission.
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When is coupling (induction) between two wires maximum?
What confirms the fact that the transfer of energy from the primary to the secondary of a transformer is not perfect?