Two resistances R & r and an inductor of inductance L are connected in series with an A.C. voltage source. Three voltmeter are also connected as shown in the figure. Reading of voltmeter, $V_{1} = 56 V, V_{2} = 120 V & V = 160 V$ Then :

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( ω L ) 2 + r 2 ​ 120 ​ = R 56 ​ = ( R + r ) 2 + ( ω L ) 2 ​ 160 ​ = i Solving R r ​ = 49 63 ​ = 7 9 ​ , ir = ( R 56 ​ ) r = 56 × 7 9 ​ = 72 volt ( B ) R 2 ω 2 L 2 ​ + ( R r ​ ) 2 = 49 225 ​ ⇒ R ω L ​ = 7 12 ​ ⇒ i ω L = 7 12 ​ × 56 = 96 Volt Power factor ( ω L ) 2 + ( R + r ) 2 ​ R ​ = 20 7 ​

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JEE Advanced (2028)