Three alternating voltage sources V1=3sinωt volt, V2=5sin(ωt+ϕ1) volt and V3=5sin(ωt−ϕ2) volt are connected across a resistance R=73Ω as shown in the figure (where ϕ1 and ϕ2 corresponds to 30° and 127° respectively). Find the rms current (in Amp) through the resistor. [ Take 2=1.4 ] (Round off to nearest integer)

v1=3sinωt;v2=5sin(ωt+ϕ1); v3=5sin(ωt−ϕ2)vmax=(532)+(1.5)2=21Imax=VmaxR=2173=21×37=3A, Irms=31.4=2.14

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Three alternating voltage sources $V_{1} = 3 \sin ω t$ volt, $V_{2} = 5 \sin (ω t + ϕ_{1})$ volt and $V_{3} = 5 \sin (ω t - ϕ_{2})$ volt are connected across a resistance $R = \sqrt{\frac{7}{3}} Ω$ as shown in the figure (where $ϕ_{1}$ and $ϕ_{2}$ corresponds to $30^{°}$ and $127^{°}$ respectively). Find the rms current (in Amp) through the resistor. [ Take $\sqrt{2} = 1.4$ ] (Round off to nearest integer)

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$v_{1} = 3 \sin ω t; v_{2} = 5 \sin (ω t + ϕ_{1}); v_{3} = 5 \sin (ω t - ϕ_{2})$

$v_{m a x} = \sqrt{(\frac{5 \sqrt{3}}{2}) + {(1.5)}^{2}} = \sqrt{21}$

$I_{\max} = \frac{V_{\max}}{R} = \frac{\sqrt{21}}{\sqrt{\frac{7}{3}}} = \sqrt{\frac{21 \times 3}{7}} = 3 A, I_{r m s} = \frac{3}{1.4} = 2.14$

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