A coil has a resistance of 10Ω and an inductance of 0.4 henry. It is connected to an AC source of 6.5 V, (30/π) Hz the average power (in watt) consumed in the circuit.

Question

A coil has a resistance of 10Ω and an inductance of 0.4 henry. It is connected to an AC source of 6.5 V, (30/π) Hz the average power (in watt) consumed in the circuit. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Average power consumed is given by P = (V2rms R/Z2) where Z = √R2 + (ω L)2 = √102 + (2π × (30/π) × 0.4)2 = 26 Ω ∴ P = (6.52 × 10/262) = 0.625 W

Q. A coil has a resistance of $10Ω$ and an inductance of $0.4$ henry. It is connected to an $A C$ source of $6.5 V, \frac{30}{π} Hz$ the average power (in watt) consumed in the circuit.

Average power consumed is given by
$P = \frac{V _{r m s}^{2} R}{Z ^{2}}$
where $Z = R^{2} + (ω L)^{2}$
$= 1 0^{2} + (2 π \times \frac{30}{π} \times 0.4)^{2}$
$= 26 Ω$
$∴ P = \frac{6. 5 ^{2} \times 10}{2 6 ^{2}} = 0.625 W$

Q. A coil has a resistance of 10Ω and an inductance of 0.4 henry. It is connected to an AC source of 6.5V,π30​Hz the average power (in watt) consumed in the circuit.

Average power consumed is given by P=Z2Vrms2​R​ where Z=R2+(ωL)2​ =102+(2π×π30​×0.4)2​ =26Ω ∴P=2626.52×10​=0.625W

Q. A coil has a resistance of $10Ω$ and an inductance of $0.4$ henry. It is connected to an $A C$ source of $6.5 V, \frac{30}{π} Hz$ the average power (in watt) consumed in the circuit.

Average power consumed is given by
$P = \frac{V _{r m s}^{2} R}{Z ^{2}}$
where $Z = R^{2} + (ω L)^{2}$
$= 1 0^{2} + (2 π \times \frac{30}{π} \times 0.4)^{2}$
$= 26 Ω$
$∴ P = \frac{6. 5 ^{2} \times 10}{2 6 ^{2}} = 0.625 W$