A non-sinusoidal voltage is v(t) = 5 + 10 * sin(2π * 50t + 30 ) 15sin(4π 0t 45 ) . o + * 5 + o This voltage is connected to the load which is a serial connection of a 10Ω resistor and a 10mH inductance. Determine: a) the power absorbed by the load, and b) Drive an expression for the load

Correct Question:

A non-sinusoidal voltage is This voltage is connected to the load which is a serial connection of a 10Ω resistor and a 10mH inductance. Determine: a) the power absorbed by the load, and b) Drive an expression for the load current

a) Power absorbed by the load = 15.1 W

b) Expression for the load current: i(t) = 0.5 + 0.95sin(2π50t + 12.57) + 1.27sin(4π50t + 12.871)

Explanation:

when V₁(t) = 5V

i₁(t) = V₁(t)/R = 5/10 = 0.5 A

When V₂(t) = 10sin(2π50t + 30)

i(t) = V(t)/(R +jX)

V₂(t) = 10∠30

i₂(t) = (10∠30)/(10 + j3.14)

i₂(t) = (10∠30)/(10.481∠14.432)

i₂(t) = 0.95∠12.47 A

For source V₃(t) = 15sin(4π50t + 45) = 15∠45

i₃(t) = V₃(t)/(R + jX)

1₃(t) = (15∠45)/(10 + j6.284)

1₃(t) = (15∠45)/(11.808∠32.129)

1₃(t) = 1.27∠12.871 A

The expression for the load current becomes:

i(t) = i₁(t) + i₂(t) + i₃(t)

i(t) = 0.5 + 0.95∠12.57 + 1.27∠12.871

i(t) = 0.5 + 0.95sin(2π50t + 12.57) + 1.27sin(4π50t + 12.871)

a) The power absorbed by the load

P = I²R

I² = 0.5² + (0.95/√2)² + (1.27/√2)²

I² = 1.51 A

P = 1.51 * 10

P = 15.1 W

answer: 500 wph.

explanation: