RL parallel circuit. No value given

[NascentOxygen]

For part E.
Branch one

R+R=2R

I=V/2R

Branch 2

R+jXL

I=V/(R+jxL)

than find the magnitude and angle than add Is suppose

That should work.

 

[Sam 007]

V*(R/(R^2+XL^2)- jXL/(R^2+XL^2)

V*√((R/(R^2+XL^2)^2- (XL/(R^2+XL^2))^2)θ= tan^-1(XL/R)Itotal = V*√((R/(R^2+XL^2)^2- (XL/(R^2+XL^2))^2) + V/2R

 

[Sam 007]

For part C... R3

V3=V*R3/(R3+jXL)
=V( R/(R+jXL)* (R-jXL)/(R-jXL)

= V*(R^2/(R^2+XL^2) - jRXL/(R^2+XL^2))mag=V*√(R^2/(R^2+XL^2))^2 - (RXL/(R^2+XL^2))^2)

θ=tan^-1(RXL/R^2)

Real power R3=(V*√(R^2/(R^2+XL^2))^2 - (RXL/(R^2+XL^2))^2))^2/R

Real power R1 and R2= V^2/2R

 

[stepfanie]

Real power R1 and R2= V/2R

isn't v^2 / 2R ?

 

[Sam 007]

isn't v^2 / 2R ?

yes ,sorry forgot to include ^2

 

[NascentOxygen]

V*(R/(R^2+XL^2)- jXL/(R^2+XL^2)

V*√((R/(R^2+XL^2)^2- (XL/(R^2+XL^2))^2)


θ= tan^-1(XL/R)


Itotal = V*√((R/(R^2+XL^2)^2- (XL/(R^2+XL^2))^2) + V/2R

You can't just place a "+" sign between the two currents and sum their magnitudes because they don't have the same angle. You can add the currents by taking their different angles into account, or, alternatively, you can separately add their real parts, and add their imaginary parts, to find the real and imaginary part of the total current.

 

[stepfanie]

thank you nascent :D you help alot..!