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A light blub is in turn connected in series (a) across an LR circuit, (b) across an RC circuit, with an AC source. Explain, giving necessary mathematical formula, the effect on the brightness of the bulb in case (a) and (b), when the frequency of the AC source is increased.

 

 

Answers (1)

Z = \sqrt{R^{2}+X_{L}^{2}}

X_{L}= j \omega L = j 2 \pi fL

X_{L}\propto frequency

As the frequency of the AC source increases X_{L} also increases. When X_{L} increases Z also increases.

As impedance increases Bulb's brightness is reduced because current flow is reduced.

(b)

Z = \sqrt{R^{2}+X_{C}^{2}}

X_{C}= \frac{1}{j\omega C}= \frac{1}{j 2\pi fC}

X_{C}\propto \frac{1}{frequency}

Thus, the frequency of the AC source increases the capacitive reactance decreases.

Therefore, the bulb glows with more brightness because more current flows through it.

Posted by

Safeer PP

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