In a series $$L-R$$ circuit, connected with a sinusoidal ac source, the maximum potential difference across L and R are $$3$$ volts and $$4$$ volts respectively.At an instant the potential difference across resistor is $$2$$ volts. The potential difference in volt, across the inductor at the same instant will be :At the same instant, the magnitude of potential difference in volt, across the source may be

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In LR CIRCUIT $$tan$$ ϕ$$=VLVR=IRMS$$ ωLIRMS $$R=34$$  As per the given $$dataVLVR=34=VL$$`$$2$$ new value of $$vL=3$$ COS $$600=1.5$$ $$VOLTV=V2L+V2R=22+1.52=2.5$$  volt

AC Circuits

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In a series L-R circuit, connected with a sinusoidal ac source, the maximum potential difference across L and R are 3 volts and 4 volts respectively.

At an instant the potential difference across resistor is 2 volts. The potential difference in volt, across the inductor at the same instant will be :

In LR CIRCUIT
tan $ϕ$ = $\frac{V_{L}}{V_{R}}$ = $\frac{I_{R M S} ω L}{I_{R M S} R}$ = $\frac{3}{4}$ As per the given data
$\frac{V_{L}}{V_{R}} = \frac{3}{4} = \frac{{V_{L}}^{`}}{2} n e w v a l u e o f v_{L} = 3 C O S 60^{0} = 1.5 V O L T$

At the same instant, the magnitude of potential difference in volt, across the source may be

V= $\sqrt{{V^{2}}_{L} + {V^{2}}_{R}} = \sqrt{2^{2} + 1 . 5^{2}} = 2.5 v o l t$

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