A resistor of 100 Ω, inductance of 1H and capacitor of
capacitance 10.13 μF are in series. The combination is
connected to an ac source of 200V, 50Hz. Find the current in
the circuit and the potential difference across the resistor.
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Explanation:
R=100 Ω . L = 1H
C = 200V , v = 50Hz , Iᴠ= ?, V = ?
Eᴠ = 200V , v = 50Hz, Iv = ?, V = ?
Xʟ = ωL = 2πvL =2 × 3.14 × 50 × 1 = 314Ω
Xᴄ = 1 / ωC = 1 / 2πvC
= 1 / 2 × 3.14 × 50 × 10.13 × 10-⁶
= 1000 / 3.14 x 10.13 = 314.38Ω ≅ 314Ω
As
Xʟ= Xᴄ
, therefore, circuit is in reasonace.
Iᴠ = Eᴠ / R = 200/ 100 2A
V = Iv x R = 2 x 100 = 200V
Hope it helps you!!
Have a nice day
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