ACL2L4

The current through the resistive branch of a parallel circuit is ? out-of-phase with the current through the inductive branch.

90 degrees 4Q7

In a parallel RL circuit, the voltage across each branch is the same.

True

In a series RL circuit, the ? are in-phase, and the ? are out-of-phase. In a parallel RL circuit, the ? are in-phase, and the ? are out-of-phase.

currents voltages voltages currents 4Q4

A circuit has an applied voltage of 240 volts and draws 5 amperes. The power factor is 70%. What is the apparent power? The values calculated for this question will be used for additional questions.

1,200 VA b/c P_app_ = E_app_ * I_T_ P_app_ = 240 * 5 P_app = 1,200 4Q15

The impedance of a parallel RL circuit is always less than the resistance or reactance of any one branch.

True 4Q12

In a parallel RL circuit, power delivered by the source is apparent power.

True 4Q20

In a parallel circuit, IR and IL must be added vectorially to find the total current.

True 4Q8

When the resistance of a parallel RL circuit is increased, the circuit becomes more inductive.

True : Note: As the resistance increases, the result is less current in the resistive branch and the reactive current becomes a greater portion of the total current. 4Q21

Regarding a parallel RL circuit where the frequency is increased: I_T_ would ?

decrease 4Q17

Regarding a parallel RL circuit where the frequency is increased: The phase angle would ?

decrease 4Q17

The power factor in a parallel RL circuit is the ratio of ? to ? .

watts / volt amps 4Q14

When the inductance of a series RL circuit increases, θ increases.

True

In a series RL circuit, power delivered by the source is apparent power.

true 4Q19

Solve for the impedance of the circuit shown by determining branch currents and total current using Ohm's Law. (Round the FINAL answer to two decimal places.) E_app_ = 6V F = 400Hz Branch 1 = R_1_ = 470 ohms Branch 2 = L = 107 mH

233.32 ohms I_R_ = E / R I_R_ = 6 / 470 I_R_ = 0.012765957 A I_L_ = E / X_L_ I_L_ = 6 / 268.78 I_L_ = 0.022322757 A I_T_ = sqrt ( I_R_^2 + I_L_^2 ) I_T_ = 0.025715271 A Z = E_T_/ I_T_ Z = 233.32 ohms

Using the formula: Z = ( R * X_L_) / sqrt( R^2 * X_L_^2 ) Solve for Z in the following: E_app_ = 6V F = 400Hz Branch 1 = R_1_ = 470 ohms Branch 2 = L = 107 mH

233.32 ohms b/c X_L_ = 2 * 3.14 * 400 * 0.107 X_L_ = 268.78 ohms Z = given wherein R = 470 and X_L_ = 268.78 4Q26

Questions missing from this reveiw

4Q1 and 11

A resistor and an inductor are connected in parallel. What is the phase relationship between the two voltages?

0 degrees (in-phase) 4Q3

The current is the same at all points in a parallel RL circuit.

False

When the resistance of a series RL circuit is increased, the circuit becomes more inductive.

False Note: Current is uniform in a series circuit; as the resistance is increased, the voltage drop across the resistor increases and it becomes a larger portion of total voltage. 4Q22

When analyzing the current in a parallel RL circuit with a large R and a small value of XL, the phase angle theta is close to 0°.

False Note: In parallel, voltage is constant; therefore, the inductive branch would have a larger current than the resistive branch and the phase angle would be closer to 90° 4Q9

True or False A resistor and an inductor are connected in parallel. The resistor draws 2 amperes and the inductance 4 amperes. The total current is 6 amperes.

False b/c The current values in parallel RL circuits cannot be added directly; the current in the resistor and the current in the inductor are 90° out-of-phase and must be added vectorially 4Q6

In a series RL circuit, the vector sum of the currents equals the total current.

False note: There is only one current in a series circuit. 4Q18

Given E = 120 V f = 60 Hz parallel branches containing -R = 30 ohms -X_L_ = 40 ohms Find I_R_ = ? I_L_ = ? I_T_ = ? Z = ? PF = ? Angle theta = ?

I_R_ = 4.00 A b/c E_R_ = E_L_ = E_R_ = 120 V I_R_ = E_R_ / R i_R_ = 120 / 30 I_R_ = 4 I_L_ =3.00 A b/c I_L_ = E_L_ / X_L_ I_L_ = 120 / 40 I_L_ = 3 I_T_ = 5.00 A I_T_ = sqrt( I^2_R_ + I^2_X_ ) I_T_ = sqrt (4^2 + 3^2) I_T_ = 5 A Z = 24 ohms b/c= E /I_T_ Z = 120 / 5 Z = 24 ohms PF = 80.0 % bc PF = I_R_ / I_T_ * 100 PF = 4 / 5 * 100 PF = 80 Angle theta = 36.9 degrees Angle theta = acos(I_R_ / I_T_ ) Angle theta = acos (4 / 5) Angle theta = 36.8 4Q10

Two resistors are connected in parallel. One has a current flow of 2 amperes and the other draws 4 amperes. What is the total current?

I_T_ = 6 A b/c I_T_ = I_R1_ + I_R2_ I_T_ = 2 + 4 I_T_ = 6 A 4Q5

Regarding a parallel RL circuit where the frequency is increased: Z would ?

Increase 4Q17

Regarding a parallel RL circuit where the frequency is increased: The power factor would ?

increase 4Q17

Regarding a parallel RL circuit where the frequency is increased: X_L_ would ?

increase 4Q17

A circuit has an applied voltage of 240 volts and draws 5 amperes. The power factor is 70%. What is the true power?

840 W b/c P_true_ = E_T_ *I_T_ *PF P_true_ = P_app_ *PF P_true_ = 1,200 * 0.7 P_true_ = 840 W 4Q16

In a parallel RL circuit, the ? can be computed by multiplying the line voltage by the line current.

apparent power 4Q13