A capacitance C whose interelectrode space is filled up with poorly conducting medium with active resistance R is connected to a source of altercating voltage $v = V_{M} \cos ω t$ .Find the maximum current $(I_{M})$ flowing in lead wires. The resistance of the wire is to be neglected.

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$I_{M} = \frac{V_{M}}{R} \sqrt{1 + 2 ω^{2} C^{2} R^{2}}$

$I_{M} = \frac{V_{M}}{R} \sqrt{1 + ω^{2} C^{2} R^{2}}$

$I_{M} = \frac{5 V_{M}}{R} \sqrt{1 + ω^{2} C^{2} R^{2}}$

$I_{M} = \frac{V_{M}}{5 R} \sqrt{1 + 2 ω^{2} C^{2} R^{2}}$

answer is B.

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Detailed Solution

If the space between electrodes is filled with poorly conducting material, then the system behaves as a parallel combination of resistance and capacitor, (shown in figure.)
Here $X_{C} = \frac{1}{ω C}$
$∴ \frac{1}{Z} = \frac{1}{R} + \frac{1}{X_{c}} (j)$ (in complex representation)
or or $\frac{1}{Z} = \frac{1}{R} + ω C j$

$∴ Z = \frac{R}{\sqrt{1 + ω^{2} C^{2} R^{2}}} ∴ i = \frac{V}{Z} = \frac{V_{M} \cos (ω t + ϕ)}{Z}$
$I = I_{M} \cos (ω t + ϕ)$
$∴ I_{M} = \frac{V_{M}}{Z} = \frac{V_{M}}{R} \sqrt{1 + ω^{2} C^{2} R^{2}}$
$\tan ϕ = \frac{R}{X_{C}} = \frac{R}{\frac{1}{ω C}} ∴ \tan ϕ = ω R C$