I need help solving this equation containing complex number and phasors
D Deleted member 4993 Guest Apr 6, 2021 #2 ParveshD.Singh said: View attachment 26228 Click to expand... Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem.
ParveshD.Singh said: View attachment 26228 Click to expand... Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem.
N nasi112 Full Member Joined Aug 23, 2020 Messages 616 Apr 6, 2021 #3 [MATH]\frac{10\angle 0^{\circ} }{2000} - \frac{V_o}{2000} + \frac{20\angle -90^{\circ} }{-j1000} - \frac{V_o}{-j1000} = \frac{V_o}{j400}[/MATH] What next?
[MATH]\frac{10\angle 0^{\circ} }{2000} - \frac{V_o}{2000} + \frac{20\angle -90^{\circ} }{-j1000} - \frac{V_o}{-j1000} = \frac{V_o}{j400}[/MATH] What next?
Steven G Elite Member Joined Dec 30, 2014 Messages 14,405 Apr 6, 2021 #4 Collect all the V0 on one side of the equation and then divide both sides by the number of V0's you have
Collect all the V0 on one side of the equation and then divide both sides by the number of V0's you have
D Deleted member 4993 Guest Apr 6, 2021 #5 nasi112 said: [MATH]\frac{10\angle 0^{\circ} }{2000} - \frac{V_o}{2000} + \frac{20\angle -90^{\circ} }{-j1000} - \frac{V_o}{-j1000} = \frac{V_o}{j400}[/MATH] What next? Click to expand... I suggest: [MATH]\frac{10\angle 0^{\circ} }{2000} +j \frac{20\angle -90^{\circ} }{1000} =\frac{V_o}{2000} - j \frac{V_o}{400} +j \frac{V_o}{1000}[/MATH] Continue.....
nasi112 said: [MATH]\frac{10\angle 0^{\circ} }{2000} - \frac{V_o}{2000} + \frac{20\angle -90^{\circ} }{-j1000} - \frac{V_o}{-j1000} = \frac{V_o}{j400}[/MATH] What next? Click to expand... I suggest: [MATH]\frac{10\angle 0^{\circ} }{2000} +j \frac{20\angle -90^{\circ} }{1000} =\frac{V_o}{2000} - j \frac{V_o}{400} +j \frac{V_o}{1000}[/MATH] Continue.....