A series LCR circuit containing an AC source | vedclass

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Question

(A) Given: $X_L = 24 \ \Omega$,$X_C = 16 \ \Omega$,$R = 6 \ \Omega$,$V = 100 \ V$.In a series $LCR$ circuit,the impedance $Z$ is given by:$Z = \sqrt{R

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) Given: $X_L = 24 \ \Omega$,$X_C = 16 \ \Omega$,$R = 6 \ \Omega$,$V = 100 \ V$.In a series $LCR$ circuit,the impedance $Z$ is given by:$Z = \sqrt{R^2 + (X_L - X_C)^2}$$Z = \sqrt{6^2 + (24 - 16)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \ \Omega$.The current $i$ in the circuit is:$i = \frac{V}{Z} = \frac{100}{10} = 10 \ A$.The potential difference across the series combination of the inductor and capacitor is given by:$V_{LC} = i |X_L - X_C|$$V_{LC} = 10 	imes |24 - 16| = 10 	imes 8 = 80 \ V$.

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