Question

d) Three equal impedances, each consisting of R and L in series are connected in star and are supplied from a 400 V, 50 Hz, 3-phase, 3-wire balanced supply mains. The input power to the load is measured by two watt-meters method and the two watt-meters read 3 kW and 1 kW. Determine the values of R and L connected in each phase. (5 Marks)

Answer 1

To determine the values of R and L connected in each phase of the circuit, we can use the two watt-meters method. The power consumed by each phase can be calculated by summing the readings of the two watt-meters. By using the power and voltage values, we can find the magnitude of the current. With the current magnitude, we can calculate the impedance and solve for the values of R and L connected in each phase.

Step-by-step explanation:

To solve this problem, we need to use the two watt-meters method to determine the values of R and L connected in each phase. Let's assume the watt-meter 1 reads W1 and the watt-meter 2 reads W2. The total power consumed by the circuit is given by the sum of the readings of the two watt-meters, which is P = W1 + W2.

Since the circuit is balanced, we can assume that the power consumed by each phase is the same. Therefore, the power consumed by each phase is P/3. We can also calculate the magnitude of the current using the formula I = P/V, where V is the line voltage.

Once we have the magnitude of the current, we can calculate the impedance using the formula Z = V/I. Since Z = R + jωL, we can equate the real and imaginary parts of the impedance to the values of R and L respectively. Finally, we can solve the equations simultaneously to find the values of R and L connected in each phase.

Answer 2

To determine the values of R and L connected in each phase in the given RLC series circuit, the two watt-meters method is used to measure the input power to the load. With the read values of 3 kW and 1 kW, the total input power is 4 kW, and each phase receives 1.33 kW. Then, using the formulas for apparent power and impedance, the values of R and L can be calculated. The values of R and L connected in each phase are approximately R = 0.198 Ω and L = 0.073 H.

Step-by-step explanation:

To determine the values of R and L connected in each phase, we can use the two watt-meters method to measure the input power to the load. Since the watt-meters read 3 kW and 1 kW, the total input power to the load is 4 kW. With a balanced 3-phase supply, each phase receives the same amount of power. Therefore, each phase receives 4/3 kW or 1.33 kW.

Assuming the power factor of the load is cos(θ), the apparent power can be calculated using the formula S = P / cos(θ), where S is the apparent power. In this case, S = 1.33 kW.

To calculate the impedance of the load, we can use the formula S = √3 * Vrms * Irms, where Vrms is the rms voltage and Irms is the rms current.

Since the load is a star-connected and balanced system, the rms voltage per phase can be calculated as Vrms = V / √3, where V is the line-to-line voltage. In this case, V = 400 V,

so Vrms = 400 / √3 ≈ 230.9 V.

Now, using the formula S = √3 * Vrms * Irms, we can rearrange it to solve for Irms.

Plugging in the values, we get 1.33 * 10^3 = √3 * 230.9 * Irms. Solving for Irms, we find that Irms ≈ 2.778 A.

Since the impedance is given by Z = Vrms / Irms, we can plug in the values to find Z ≈ 230.9 / 2.778 ≈ 83.08 Ω.

Given that the impedance consists of R and L in series, we can express it as Z = √(R^2 + (ωL)^2), where ω is the angular frequency and L is the inductance. In this case, ω = 2π * 50

= 314 rad/s.

Simplifying the equation, we get 83.08 = √(R^2 + (314 * L)^2).

Squaring both sides and rearranging the equation, we find (R^2 + (314 * L)^2) = 83.08^2.

To continue solving for R and L, we need another equation. We can use the phase difference between the current and the voltage at the load.

For a balanced 3-phase system, the power factor for a star-connected and inductive load is cos(θ) = R / Z. In this case, cos(θ) = R / 83.08.

Since we have three equal impedances connected in star, we can assume that the phase difference is the same for all three.

Given that the two watt-meters measure the power for a single phase, the power factor can also be calculated as cos(θ) = P / (√3 * Vrms * Irms), where P is the power measured by the watt-meters.

Plugging in the values for P, Vrms, and Irms, we get cos(θ) = 1 / (3 * 230.9 * 2.778) ≈ 0.00238.

Setting the two expressions for cos(θ) equal to each other, we have R / 83.08 ≈ 0.00238. Solving for R, we find that R ≈ 0.00238 * 83.08 ≈ 0.198 Ω.

Now, substituting the value of R into the equation (R^2 + (314 * L)^2) = 83.08^2,

we have (0.198^2 + (314 * L)^2) = 83.08^2.

Simplifying and solving for L, we find L ≈ √(83.08^2 - 0.198^2) / 314 ≈ 0.073 H.

Therefore, the values of R and L connected in each phase are approximately R = 0.198 Ω and L = 0.073 H.