A capacitor of capacitance 100- F and the coil of resistance 50- and inductance 0-5 H are connected in series with a 110 V- 50Hz AC source- Find the rms value of the current-

The correct option is A 0.82A [ ω #160; = 2π f = 2π #160; × 50 = ...

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Standard XII

Physics

Displacement Current

A capacitor o...

Question

A capacitor of capacitance $100 μ F$ and the coil of resistance $50 Ω$ and inductance 0.5 H are connected in series with a 110 V, 50Hz AC source. Find the rms value of the current.

0.12A

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0.82A

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0.53A

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0.22A

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Solution

The correct option is A 0.82A
$\begin{matrix} ω = 2 π f = 2 π \times 50 = 100 π = 314 r a d s^{- 1} X c = \frac{1}{ω c} = \frac{1}{314 \times 100 \times 10^{- 6}} = 31.84 Ω X_{L} = ω L = 314 \times 0.5 = 157 Ω Z = \sqrt{R^{2} {(X_{L} - X_{c})}^{2}} = \sqrt{50^{2} {(157 - 31.84)}^{2}} = 134.77 Ω \end{matrix}$
$I_{r m s} = \frac{V_{r m s}}{Z} = \frac{100}{134.77} = 0.814 A = 0.82 A (a p p r o x)$

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A capacitor of capacitance 100μF and the coil of resistance 50Ω and inductance 0.5 H are connected in series with a 110 V, 50Hz AC source. Find the rms value of the current.

The correct option is A 0.82Aω=2πf=2π×50=100π=314rads−1Xc=1ωc=1314×100×10−6=31.84ΩXL=ωL=314×0.5=157ΩZ=√R2(XL−Xc)2=√502(157−31.84)2=134.77ΩIrms=VrmsZ=100134.77=0.814A=0.82A(approx)

A capacitor of capacitance $100 μ F$ and the coil of resistance $50 Ω$ and inductance 0.5 H are connected in series with a 110 V, 50Hz AC source. Find the rms value of the current.