A resistor of 200 Ω and a capacitor of 15 μ F are connected in series to a 220 V , 50 Hz AC source. The current in the circuit is

Question

A resistor of 200 Ω and a capacitor of 15 μ F are connected in series to a 220 V , 50 Hz AC source. The current in the circuit is (A) 755 A (B) 7.55 mA (C) 0.755 A (D) 0.755 mA

Tardigrade Admin · Accepted Answer

Impedance of the circuit Z =√R2+XC2=√R2+(2 π v C)-2 =√(200 Ω)2+(2 × 3.14 × 50 × 15 × 10-6 F )-2 =√(200 Ω)2+(212 Ω)2=291.5 Ω Therefore, the current in the circuit is i=V / Z=(220 V /291.5 Ω)=0.755 A

Q. A resistor of $200 Ω$ and a capacitor of $15 μ F$ are connected in series to a $220 V, 50 Hz$ AC source. The current in the circuit is

755 A

7.55 mA

0.755 A

0.755 mA

Impedance of the circuit
$Z = R^{2} + X_{C}^{2} = R^{2} + (2 π v C)^{- 2}$
$= (200 Ω)^{2} + (2 \times 3.14 \times 50 \times 15 \times 1 0^{- 6} F)^{- 2}$
$= (200 Ω)^{2} + (212 Ω)^{2} = 291.5 Ω$
Therefore, the current in the circuit is
$i = V / Z = \frac{220 V}{291.5 Ω} = 0.755 A$

Q. A resistor of 200Ω and a capacitor of 15μF are connected in series to a 220V,50Hz AC source. The current in the circuit is