Test Index

Alternating Current

Section: Physics

Question : 9 of 13

Marks: +1, -0

Solution:

From the given circuit, we have

I_{\text{capacitance}}

\frac{20}{100\sqrt{2}}

(progress ahead of the voltage by + π/4)

I_{\text{inductive}}

\frac{20}{50\sqrt{2}}

(lagging behind the voltage by -π/4)

Therefore,

I =

\sqrt{I_1^2+I_2^2}

\sqrt{\frac{1}{10}}

~ 0.3

V_{100\Omega}

\frac{20}{100\sqrt{2}} \times 100

10\sqrt{2}\,\text{V}

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