A series RL circuit contains two resistors and two inductors. The resistors are 27 W and 47 W. The inductors have inductive reactances of 50 W and 40 W. The applied voltage is 120 V. What is the voltage drop on the inductor that has 40 W of reactance?
Given:
E = 120 volts.
R1 = 27 Ohms.
R2 = 47 Ohms.
X1 = j50 Ohms.
X2 = j40 Ohms.
,
Z = (R1+R2) + j(X1+X2) = Total impedance.
Z = 74 + j90 = 116.5 Ohms[50.6o].
Z = Sqrt(74^2+90^2.
Tan A = 90/74.
I = E/Z = 120[0o]/116.5[50.6o] = 1.03Amps[-50.6o].
The negative angle means that the current lags the applied voltage by 50.6o.
V2 = I * X2 = 1.03[-50.6] * 40[90o] = 41.2 Volts & 39.4o.
To calculate the voltage drop across the inductor with 40 Ω of reactance in a series RL circuit, we need to find the total impedance of the circuit first. The impedance of an RL circuit is given by:
Z = √(R^2 + X_L^2)
Where:
R is the resistance of the circuit
X_L is the inductive reactance
Given data:
Resistance of the resistors, R1 = 27 Ω
Resistance of the resistors, R2 = 47 Ω
Inductive reactance of the inductor, X_L1 = 50 Ω
Inductive reactance of the inductor, X_L2 = 40 Ω
To find the total impedance, we can add the resistance and inductive reactance of each component:
Z_total = Z_R1 + Z_R2 + Z_L1 + Z_L2
Where:
Z_R1 = R1
Z_R2 = R2
Z_L1 = X_L1 (since there is no resistance in the inductor)
Z_L2 = X_L2 (since there is no resistance in the inductor)
Substituting the given values:
Z_total = 27 + 47 + 50 + 40
Z_total = 164 Ω
Now, we can calculate the voltage drop across the inductor with 40 Ω of reactance using Ohm's law:
V_L2 = I * X_L2
Where:
V_L2 is the voltage drop across the inductor
I is the current flowing through the circuit
To find the current, we can use the formula:
I = V_total / Z_total
Substituting the given values:
V_total = 120 V (applied voltage)
Z_total = 164 Ω (total impedance)
I = 120 / 164
I ≈ 0.73 A
Now, substituting the values of I and X_L2, we can calculate the voltage drop:
V_L2 = 0.73 * 40
V_L2 ≈ 29.2 V
Therefore, the voltage drop across the inductor with 40 Ω of reactance is approximately 29.2 V.
To find the voltage drop across the inductor with 40Ω of reactance in the series RL circuit, we need to calculate the total impedance of the circuit first and then use Ohm's Law to determine the voltage drop.
1. Calculate the total impedance (Z) of the circuit:
The impedance in a series RL circuit is given by the formula Z = √(R^2 + (XL - XC)^2), where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance (which is zero in this case, as it's an RL circuit).
Z = √((27Ω)^2 + (40Ω)^2)
= √(729Ω + 1600Ω)
= √(2329Ω)
≈ 48.25Ω
2. Apply Ohm's Law to determine the voltage drop across the inductor with 40Ω of reactance:
Ohm's Law states that V = I * Z, where V is the voltage drop, I is the current flowing through the circuit, and Z is the impedance.
Since the circuit is a series circuit, the total current (I) is the same throughout the circuit.
I can be calculated using Ohm's Law V = I * R, where V is the applied voltage and R is the resistance.
I = V / R
= 120V / 27Ω
≈ 4.44A
Now, we can calculate the voltage drop (V) across the inductor with 40Ω of reactance using Ohm's Law:
V = I * Z
= 4.44A * 48.25Ω
≈ 213.94V
Therefore, the voltage drop across the inductor that has 40Ω of reactance is approximately 213.94 V.