Question

An inductor of inductance
$\bigg[\rm L=\dfrac{1}{100\pi}\:H\bigg][L=100π1H]$

a capacitor of capacitance

$\bigg[\rm C=\dfrac{1}{500\pi}\:H\bigg][C= 500π1​H]$

and a resistor of resistance 2 Ω are connected in series with an AC voltage source as shown in the figure.

The voltage of the AC source is given as V = 10cos(100π t)volt.
What will be the potential difference between A and B ?​

Answer 1

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Answer 2

V = 10 cos (100лt)}w - 100л total impedance of circuit

z = √(ac - XL)² + R² ac 1 WC 5 a₁ = wL=1

Z = √(51)² + 3² : = 5 Phase difference o = tan-¹ Xc - XL R

5-1

3

= tan-1

= 53°

Current V 10 5 = = cos (100лt 53°)

i = 2 cos (100лt – 53°)

impedance between AB

(Xc - XL) = 4

Potential difference between AB

= iR = 2 cos (100лt - 53) × 4

= 8 cos(100лt - 53) volt

= option B

Answer is 8 cos(100лt - 53) volt