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An ideal transformer has 1500 turns and 75 turns in primary and Bartleby Bartleby Bartleby Bartleby Bartleby Bartleby Bartleby Bartleby Bartleby Ba secondary coils respectively. If 10.4 amps flows in the primary side. Barueuy bardieuy Darley DarLieu DarLevy Darlieuy Darley DarLieby Darlieuy Da what is the load current? Heby Bartleby Bartleby Bartleby Bartleby Bartleby Ba

An ideal transformer has 1500 turns and 75 turns in primary and Bartleby Bartleby Bartleby Bartleby Bartleby Bartleby Bartleby Bartleby Bartleby Ba secondary coils respectively. If 10.4 amps flows in the primary side. Barueuy bardieuy Darley DarLieu DarLevy Darlieuy Darley DarLieby Darlieuy Da what is the load current? Heby Bartleby Bartleby Bartleby Bartleby Bartleby Ba

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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**Transformer Calculation** An ideal transformer has 1500 turns and 75 turns in primary and secondary coils respectively. If 10.4 amps flows in the primary side, what is the load current? --- To solve for the load current in an ideal transformer, we can use the formula: \[ \frac{I_p}{I_s} = \frac{N_s}{N_p} \] Where: - \( I_p \) = primary current - \( I_s \) = secondary current (load current) - \( N_p \) = number of turns in the primary coil - \( N_s \) = number of turns in the secondary coil Given: - \( I_p = 10.4 \) amps - \( N_p = 1500 \) turns - \( N_s = 75 \) turns Substitute the known values into the formula to find \( I_s \): \[ \frac{10.4}{I_s} = \frac{75}{1500} \] Solving for \( I_s \): \[ I_s = \frac{10.4 \times 1500}{75} \] Calculate \( I_s \): \[ I_s = 208 \] amps Thus, the load current is 208 amps.
Transcribed Image Text:**Transformer Calculation** An ideal transformer has 1500 turns and 75 turns in primary and secondary coils respectively. If 10.4 amps flows in the primary side, what is the load current? --- To solve for the load current in an ideal transformer, we can use the formula: \[ \frac{I_p}{I_s} = \frac{N_s}{N_p} \] Where: - \( I_p \) = primary current - \( I_s \) = secondary current (load current) - \( N_p \) = number of turns in the primary coil - \( N_s \) = number of turns in the secondary coil Given: - \( I_p = 10.4 \) amps - \( N_p = 1500 \) turns - \( N_s = 75 \) turns Substitute the known values into the formula to find \( I_s \): \[ \frac{10.4}{I_s} = \frac{75}{1500} \] Solving for \( I_s \): \[ I_s = \frac{10.4 \times 1500}{75} \] Calculate \( I_s \): \[ I_s = 208 \] amps Thus, the load current is 208 amps.
The nameplate on a single-phase, step-down transformer indicates that it’s rated at 33.33 KVA, 7967V-120V. Find the rated currents at the primary and secondary sides.
Transcribed Image Text:The nameplate on a single-phase, step-down transformer indicates that it’s rated at 33.33 KVA, 7967V-120V. Find the rated currents at the primary and secondary sides.
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