An ac source of 50v,100÷πHz is connected in
series with an inductor of inductance L ad a
resistor of resistance R.the current in the circuit
is 2A ad p.d. across L and R are 30v ad 40v
respectively
1) calc d average power dissipated 2) calc d power factor in the circuit
1. The resistance of the inductor is assumed to be zero.
P = Vr*I = 40 * 2 = 80 Watts.
2. pf = Cos A = Vr/E = 40/50 = 0.8.
To calculate the average power dissipated and the power factor in this circuit, we need to use the formulas associated with AC circuits.
1) Calculation of Average Power Dissipated:
The average power dissipated can be calculated using the formula: P_avg = I_avg * V_avg * cos(θ), where P_avg is the average power, I_avg is the average current, V_avg is the average voltage, and cos(θ) is the power factor.
To find the average current and average voltage, we need to convert the given values (peak current and peak voltage) to their respective average values.
Given:
AC source peak voltage (Vp) = 50V
AC source frequency (f) = 100/π Hz
Peak current (Ip) = 2A
P.d. across inductor (VL) = 30V
P.d. across resistor (VR) = 40V
Average current (I_avg) = Ip / √2
= 2A / √2
= 2√2 A ≈ 2.83 A
Average voltage (V_avg) = Vp / √2
= 50V / √2
= 50√2 V ≈ 35.36 V
Now we can calculate the average power dissipated:
P_avg = I_avg * V_avg * cos(θ)
To find the power factor (cos(θ)), we need to calculate the phase difference (θ) between the current and voltage. Since we know the voltage across the inductor (VL) and resistor (VR), we can use the arctan formula to find the phase difference:
θ = arctan(VR / VL)
θ = arctan(40V / 30V)
θ = arctan(4/3)
θ ≈ 53.13°
Power factor (cos(θ)) = cos(53.13°)
Power factor (cos(θ)) ≈ 0.602
Therefore, the average power dissipated in the circuit (P_avg) can be calculated by substituting the values:
P_avg = I_avg * V_avg * cos(θ)
P_avg = 2.83 A * 35.36 V * 0.602
P_avg ≈ 60.19 Watts
2) Calculation of Power Factor:
As calculated earlier, the power factor (cos(θ)) is approximately 0.602, indicating that the circuit is inductive. It means the current lags behind the voltage by approximately 53.13°.
Hence, the power factor in the circuit is approximately 0.602.