Lecture 21, ACT 3
The primary coil of an ideal transformer is connected to an AC voltage source as shown. There are 50 turns in the primary and 200 turns in the secondary.
- If V1 = 120 V, what is the potential drop across the resistor R ?
The ratio of turns, (N2/N1) = (200/50) = 4 The ratio (V2/V1) = (N2/N1). Therefore, (V2/V1) = 480 V
- If 960 W are dissipated in the resistor R, what is the current in the primary ?
Gee, we didn’t talk about power yet….
But, let’s assume energy is conserved…since it usually is around here
Therefore, 960 W should be produced in the primary
P1 = V1 I1 implies that 960W/120V = 8 A