ACL2L9
Is the total parallel circuit inductive or capacitive in the circuit shown? Given E = 48 V Branch 1 R_1_ = 60 ohms Branch 2 X_L_ = 40 ohms R_2_ = 30 ohms
Capacitve b/c Current through the 50-ohm capacitive branch is at 90° leading. Current through the 50-ohm RL branch is at 53.13° lagging. Therefore, the overall current for the parallel component results in more lead current than lag current. 9Q15
In the combination circuit shown, solve first for the parallel impedance (Z) assuming a voltage of 10 V. Next, solve for the total circuit impedane. Last, solve for the individual currents, voltages and the phase angle. Given E = 15 V f = 50 Hz series R_1_ = 10 ohms Branched X_C_ = 10 ohms R_2_ = 10 ohms Find. Z_R2,C_ Z_total_ I_T_ I_C_ I_R1_ I_R2_ E_R1_ E_R2_ E_C_ Angle theta
E_ASSUMED = 10 I_XC-A_ = E_ASSUMED_ / X_C_ I_XC-A_ = 10 / 10 I_XC-A_ = 1 I_T-A_ = sqrt(I_R-A_^2 +I_XC-A_^2) I_T-A_ = sqrt(1 + 1) I_T-A_ = 1.414213562373095 Z_R2-C_ = E_ASSUMED_ / I_T-A Z_R2-C_ = 10 / 1.41 Z_R2-C_ = 7.07 ohms A_NGLE-R2C_ = acos (I_R-A/ I_T-A) A_NGLE-R2C_ = acos (1 / 1.41) A_NGLE-R2C_ = 50 H_C_ = Z_R2-C_ 8 cos(A_NGLE-R2c) H_C_ = 7.07 *cos(45) H_C_ = 5. V_C_ = Z_R2-c * sin (A_ngle-r2C) V_C_ = 7.07 * sin (45) V_C_ = 5 H_TOT_ = R_1_ + H_C_ H_TOT_ = R_1 +H_C_ H_TOT_ = 10 + 5 H_TOT_ = 14 Z_total_ = sqrt(H_TOT_^2 + V_C_^2) Z_total_ = sqrt(15^2 + 5^2) Z_total_ = 15.81 ohms Angle theta = atan(V_C_ / H_TOT_) Angle theta = atan(5 / 15) Angle theta = 18.43 degrees I_T_ = E / Z_total_ I_T_ = 15 / 15.81 I_T_= 949mA I_R1_ = I_T_ = 949 mA E_C_ = I_T_ * Z_R2-C E_C_ = 6.71 V E_R2_ = E_C_ = 6.71 V E_R1_ = I_R1_ * R_1_ E_R1_ = .9486832980505140*10 E_R1_ = 9.48683298050514 I_R2_ = E_R2_ / R_2_ I_R2_ = 6.71/10 I_R2_ = .671 I_C_ = E_C_ / X_C_ I_C_ = 6.71 / 10 I_C_ = .670820393249937 91q11
True or False To make the circuit shown inductive, the frequency would have to be increased. Given E = 48 V Series R_1_ - 60 ohms Branch 1 X_C_ = 50 ohms Branch 2 X_L_ - 40 ohms R_2_ = 30 ohms
False 9Q18
Given E = 48 V Series R_1_ - 60 ohms Branch 1 X_C_ = 50 ohms Branch 2 X_L_ - 40 ohms R_2_ = 30 ohms The parallel circuit cannot have a greater impedance value than the 50 ohms in the capacitive branch.
False Note: The total current needed for the branches is reduced due to the circulating current between the inductor and the capacitor, which causes the inductive current to cancel the capacitive current. 9Q16
Determine the impedance of the parallel RC branch. (Assume 10 volts across the branch.) The values calculated for this question may be used for additional questions. (Round the FINAL answer to two decimal places.) E_app_ = 15 V f = 50 Hz series = R = 10 ohm Br1 = X_C_ = 10 ohms Br2 = R = 10 ohms
I_C_ = E_C_ / X_C_ I_C_ = 10 / 10 I_C_= 1 A I_R_ = E_R_ / R I_R_ = 10 / 10 I_R_ = 1 A I_T_ = sqrt (I_R_^2 + I_C_^2) I_T_ = sqrt (1 + 1) I_T_ = 1.414 A Z_RC_ = E_RC_ / I_T_ Z_RC_ = 10 / 1.414 Z_RC_ = 7.07 ohms 9Q8
Determine the impedance of the parallel RC branch. (Assume 16 volts across the branch.) The values calculated for this question may be used for additional questions. (Round the FINAL answer to two decimal places.) E_app_ = 15 V f = 50 Hz series = R = 10 ohm Br1 = X_C_ = 10 ohms Br2 = R = 10 ohms
I_C_ = E_C_ / X_C_ I_C_ = 16 / 10 I_C_= 1.6 A I_R_ = E_R_ / R I_R_ = 16 / 10 I_R_ = 1.6 A I_T_ = sqrt (I_R_^2 + I_C_^2) I_T_ = sqrt (1.6^2 + 1.6^2) I_T_ = 2.263 A Z_RC_ = E_RC_ / I_T_ Z_RC_ = 16 / 2.263 Z_RC_ = 7.07 ohms 9Q9
If the impedance of the parallel RC branch in circuit shown is calculated using different assumed voltage values, the value of impedance will remain the same. E_app_ = 15 V f = 50 Hz series = R = 10 ohm Br1 = X_C_ = 10 ohms Br2 = R = 10 ohms
True 9Q10
A vector times the sine of theta will give the vertical component. The horizontal component can be determined by multiplying the vector by the cosine of theta.
True 9Q6
In the combination circuit shown, solve first for the impedance of the inductive branch and then for the parallel impedance (Z). Next, solve for the toatal circuit impedance. Last solve for the final values shown including the final circuit phase angle. Given E = 48 V Series R_1+ 60 ohms branched X_C_ = 50 ohms = 40 ohms Branch #2 R_2_ = 30 ohms X_L_ = 40 ohms
Z_R2-L_ = sqrt(R_2_^2 + X_L_^2) Z_R2-L_ = sqrt(30^2 + 40^2) Z_R2-L_ = 50_ A_NGLE-R2L_ = atan (X_L_ / R^2) A_NGLE-R2l_ = atan (40 / 30) A_NGLE-R2L_ = 53.13010235415597 E_ASSUMED_ = 50 I_C-A_ = E_ASSUMED_ / X_C_ I_C-A_ = 50 / 50 I_C-A_ = 1 I_LR-A_ = E_ASSUMED / Z_R2-L_ I_LR-A_ = 50 / 50 I_LR-A_ = 1 H_C-A_ = I_LR-A_ * cos(A_ngle-R2L_) H_C-A_ = 1 * cos (53.13 H_C-A_ = .6 V_C-A_ = I_R-A_ * sin(A_NGLE-R2L_) V_C-A_ = 1 * sin(53.13) V_C-A_ = .8 V_TOT-R2LC_ = I_C-A_ - V_C-A V_TOT-R2LC_ = 1.00 - .8 V_TOT-R2LC_ = .2 H_TOT-R2LC_ = H_C-A_ = .6 Z_R2-L-C_ = E_ASSUMED / I_TOT-ASSUMEC Z_R2-L-C = 50 /.6324555320336760 Z_R2-L-C_ = 79.05694150420946 **** THIS TAKE A ****ING SCREENSHOT OF THIS GODFORSAKEN SHIT**** OF A ANSWER. LJKADSFUKGIAHRKLJHLK;AERHOI;JAOJLHGKLR;J NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
Capacitive reactance is determined by the ? and the ? .
capacitance / frequency 9Q5
In a capacitive circuit, the ? leads the ? .
current / voltage 9Q3
E_app_ = 15 V f = 50 Hz series = R = 10 ohm Br1 = X_C_ = 10 ohms Br2 = R = 10 ohms In the circuit shown, current will ? the voltage.
lead 9Q12
Given E = 48 V Series R_1_ - 60 ohms Branch 1 X_C_ = 50 ohms Branch 2 X_L_ - 40 ohms R_2_ = 30 ohms The line current ? the line voltage in the circuit shown.
leads 9Q17
The reference for analyzing any ? circuit is the voltage.
parallel 9Q2
The reference for analyzing any ? circuit is the current.
series 9Q1
In an inductive circuit, the ? leads the ? .
voltage / current 9Q4
