In an AC circuit, a capacitor of capacitance C=(25/π) μ F and a resistor of resistance R=300 Ω are connected in series with an AC source of 200 V and 50 s -1 frequency. The power dissipated (in watt) in the circuit will be

Question

In an AC circuit, a capacitor of capacitance C=(25/π) μ F and a resistor of resistance R=300 Ω are connected in series with an AC source of 200 V and 50 s -1 frequency. The power dissipated (in watt) in the circuit will be (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

cos φ=(R/Z)=(R/√XC2+R2)=(300/500)=0.6 P=V textrms i text rms cos φ =(V textrms 2/z) cos φ=((200)2/500) × 0.6=48 W

Q. In an $A C$ circuit, a capacitor of capacitance $C = \frac{25}{π} μ F$ and a resistor of resistance $R = 300 Ω$ are connected in series with an AC source of $200 V$ and $50 s^{- 1}$ frequency. The power dissipated (in watt) in the circuit will be

$cos ϕ = \frac{R}{Z} = \frac{R}{X _{C}^{2} + R ^{2}} = \frac{300}{500} = 0.6$
$P = V_{rms} i_{rms} cos ϕ$
$= \frac{V _{rms}^{2}}{z} cos ϕ = \frac{( 200 ) ^{2}}{500} \times 0.6 = 48 W$

Q. In an AC circuit, a capacitor of capacitance C=π25​μF and a resistor of resistance R=300Ω are connected in series with an AC source of 200V and 50s−1 frequency. The power dissipated (in watt) in the circuit will be

cosϕ=ZR​=XC2​+R2​R​=500300​=0.6 P=Vrms ​i rms ​cosϕ =zVrms2​​cosϕ=500(200)2​×0.6=48W

Q. In an $A C$ circuit, a capacitor of capacitance $C = \frac{25}{π} μ F$ and a resistor of resistance $R = 300 Ω$ are connected in series with an AC source of $200 V$ and $50 s^{- 1}$ frequency. The power dissipated (in watt) in the circuit will be

$cos ϕ = \frac{R}{Z} = \frac{R}{X _{C}^{2} + R ^{2}} = \frac{300}{500} = 0.6$
$P = V_{rms} i_{rms} cos ϕ$
$= \frac{V _{rms}^{2}}{z} cos ϕ = \frac{( 200 ) ^{2}}{500} \times 0.6 = 48 W$