(a) To find the value of the load resistance, we need to consider the transformer ratio. The secondary voltage will be equal to the primary voltage divided by the turns ratio. In this case, the turns ratio is 16:1, so the secondary voltage will be 20 V / 16 = 1.25 V.
Next, we need to calculate the total impedance of the load. The load impedance will be the parallel combination of the resistance of the load and the impedance of the transformer. Since the load resistance is equal to the resistance of the transformer, we can calculate the total impedance as:
Z_load = R_load || R_transformer
Z_load = (R_load * R_transformer) / (R_load + R_transformer)
Z_load = (20kΩ * 20kΩ) / (20kΩ + 20kΩ)
Z_load = 10kΩ
Therefore, the value of the load resistance is 10kΩ.
(b) To calculate the power dissipated in the load, we can use the formula:
P = V^2 / R
where V is the voltage across the load (1.25 V) and R is the load resistance (10kΩ). Plugging in the values, we get:
P = (1.25 V)^2 / 10kΩ
P = 1.5625 / 10000
P = 0.00015625 W
Therefore, the power dissipated in the load is 0.00015625 W, or 156.25 μW.
An a.c. source of 20 V and internal resistance 20 k Ω is matched to a load by a
16:1 single-phase transformer. Determine (a) the value of the load resistance
and (b) the power dissipated in the load
1 answer