For the circuit in Fig. 9.50, use a phasor diagram to find the value of R that will cause the current through that resistor, iR , to lag the source current, is ,by 45° when ω=5krad/s.

Question

For the circuit in Fig. 9.50, use a phasor diagram to find the value of R that will cause the current through that resistor,[latex] i_{R}[/latex], to lag the source current,[latex] i_{s}[/latex],by 45° when [latex]\omega=5krad/s[/latex].

Accepted Answer

By Kirchhoff’s current law, the sum of the currents [latex]\pmb{I}_{R},\pmb{I}_{L},[/latex]and[latex]\pmb{I}_{C}[/latex] must equal the source current[latex]\pmb{I}_{s}[/latex]. If we assume that the phase angle of the voltage [latex]\pmb{V}_{m}[/latex] is zero, we can draw the current phasors for each of the components. The current phasor for the inductor is given by

[latex]\pmb{I}_{L} =\frac{\pmb {V}_{m}\angle 0^{\circ}}{j\left(5000 ight)\left(0.2 imes 10^{-3} ight)}= \pmb {V}_{m} \angle -90^{\circ}[/latex],

whereas the current phasor for the capacitor is given by

[latex]\pmb{I}_{C} =\frac{\pmb {V}_{m}\angle 0^{\circ}}{-\frac{j}{\left(5000 ight)\left(800 imes 10^{-6} ight)}}= 4\pmb {V}_{m} \angle 90^{\circ}[/latex],

and the current phasor for the resistor is given by

[latex]\pmb{I}_{R} =\frac{\pmb {V}_{m}\angle 0^{\circ}}{R}= \frac{\pmb {V}_{m}}{R}\angle 0^{\circ}[/latex].

These phasors are shown in Fig. 9.51. The phasor diagram also shows the source current phasor, sketched as a dotted line, which must be the sum of the current phasors of the three circuit components and must be at an angle that is 45° more positive than the current phasor for the resistor. As you can see, summing the phasors makes an isosceles triangle, so the length of the current phasor for the resistor must equal [latex]3\pmb{V}_{m}[/latex] Therefore, the value of the resistor is [latex] \frac{1}{3}\Omega [/latex].

Question 9.15: For the circuit in Fig. 9.50, use a phasor diagram to find t...

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