A 300 Ω resistor and a capacitor of (25/) μF are connected in series to a 200V- 50Hz ac source. The current in the circuit is –

R=300Ω,C=25π×10−6F Impedance, Z=R2+XC2=(300)2+12π×50×25π×10−62=(300)2+(400)2=500Iras =ErmssZ=200500=0.4A

A 300 Ω resistor and a capacitor of (25/) μF are connected in series to a 200V- 50Hz ac source. The current in the circuit is –

R=300Ω,C=25π×10−6F Impedance, Z=R2+XC2=(300)2+12π×50×25π×10−62=(300)2+(400)2=500Iras =ErmssZ=200500=0.4A

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A 300 Ω resistor and a capacitor of (25/𝜋) μF are connected in series to a 200V- 50Hz ac source. The current in the circuit is –

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0.6 A

0.1 A

0.4 A

0.8 A

answer is B.

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Detailed Solution

$R = 300 Ω, C = (\frac{25}{π} \times 10^{- 6}) F Impedance, Z = \sqrt{R^{2} + X_{C}^{2}}$
$\begin{array}{l} = \sqrt{(300)^{2} + {(\frac{1}{2 π \times 50 \times \frac{25}{π} \times 10^{- 6}})}^{2}} \\ = \sqrt{(300)^{2} + (400)^{2}} \\ = 500 \\ I_{ras} = \frac{E_{rmss}}{Z} = \frac{200}{500} = 0 .4 A \end{array}$