A coil takes a current of 15A and dissipates 2800W when connected with the supply of 200V ,50Hz when another coil is connected in parallel with it , it takes 30 A at a power factor of 0.866 lagging . Determine the current and power power when it connected in series
To determine the current and power when the coil is connected in series, we need to understand the concept of series RL circuits in AC power systems.
In a series RL circuit, a resistor (R) and an inductor (L) are connected in series with a power source. The behavior of the circuit is influenced by the resistance and inductance values, as well as the frequency of the power source.
In this scenario, the given coil dissipates 2800W when connected to a supply of 200V and 50Hz, with a current of 15A. We can use this information to calculate the values of resistance (R) and inductance (L) for the coil.
To calculate the resistance (R), we can use the formula:
Power (P) = Current (I) * Voltage (V) * Power Factor (PF), where PF = 1 for a purely resistive load.
So, 2800W = 15A * 200V * 1.
Rearranging the equation, we get:
R = Power / (Current * Voltage) = 2800W / (15A * 200V) = 0.933Ω.
Now, to calculate the inductance (L), we can use the formula for the inductive reactance (XL), which is a measure of the opposition to current flow due to inductance:
XL = Voltage (V) / Current (I).
In this case, XL = 200V / 15A = 13.33Ω.
Now, when another coil is connected in parallel with the existing coil, the equivalent circuit will have a new resistance and inductance value. However, the supply voltage remains the same at 200V.
The new current, when the coil is connected in parallel, is given as 30A at a power factor of 0.866 lagging. To calculate the power consumed (P), we can use the formula:
P = Current (I) * Voltage (V) * Power Factor (PF).
So, P = 30A * 200V * 0.866 = 5190W.
When the coil is connected in series, the total current passing through the circuit will be different. To determine the new current, we can use the following steps:
1. Calculate the total impedance (Z) of the series RL circuit using the formula:
Z = sqrt(R² + (XL - XC)²), where XC is the capacitive reactance (which will be zero in this case because there is no capacitor).
In this case, Z = sqrt((0.933Ω)² + (13.33Ω)²) = 13.419Ω.
2. Calculate the total current (I_total) using Ohm's Law:
I_total = Voltage (V) / Impedance (Z) = 200V / 13.419Ω = 14.91A (approximately).
3. Calculate the power (P) when the coil is connected in series using the formula:
P = Current (I_total) * Voltage (V) * Power Factor (PF).
So, P = 14.91A * 200V * 0.866 = 2567.92W (approximately).
Therefore, when the coil is connected in series, the current passing through the circuit will be approximately 14.91A, and the power consumed will be approximately 2567.92W.