An inductor ( x L = 2Ω ) a capacitor ( x C = 8Ω ) an a resistance ( 8Ω ) is connected in series with an A.C. source. The voltage output of A.C. source is given by v = 10 cos 100 π t (volts). If th

Question

An inductor ( x L ​ = 2Ω ) a capacitor ( x C ​ = 8Ω ) an a resistance ( 8Ω ) is connected in series with an A.C. source. The voltage output of A.C. source is given by v = 10 cos 100 π t (volts). If the instantaneous potential difference between A and B is x when it is half of the voltage output from source at that instant than what is the value of x (in Volts) :

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A-B spans L and C, so by KVL around the source side: $v_{A B} = v_{s} - v_{R}$ .

Condition: $v_{A B} = \frac{v _{s}}{2}$ , hence $v_{R} = \frac{v _{s}}{2}$ at that instant.

$Z = 8^{2} + (2 - 8)^{2} = 10Ω$ , so $I_{0} = \frac{10}{10} = 1 A$ ; net capacitive, current leads v by $ϕ = tan^{- 1} \frac{6}{8} = 3 7^{\circ}$ .

Take $v_{s} = 10 cos θ$ ; then $v_{R} = 8 cos (θ + 3 7^{\circ})$ .

$8 cos (θ + 3 7^{\circ}) = 5 cos θ$ gives $6.4 cos θ - 4.8 sin θ = 5 cos θ$ , so $tan θ = \frac{1.4}{4.8} = 0.29$ , $cos θ = 0.96$ .

$x = \frac{v _{s}}{2} = 5 cos θ = 5 \times 0.96 = 4.80 V$ .

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