If a resistor of resistance 4 Omega,a capacit | vedclass

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(D) In an $LCR$ series circuit,the impedance $Z$ is given by the formula:$Z = \sqrt{R^2 + (X_L - X_C)^2}$Given values are:Resistance $R = 4 \ \Omega$C

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$5$

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(D) In an $LCR$ series circuit,the impedance $Z$ is given by the formula:$Z = \sqrt{R^2 + (X_L - X_C)^2}$Given values are:Resistance $R = 4 \ \Omega$Capacitive reactance $X_C = 6 \ \Omega$Inductive reactance $X_L = 9 \ \Omega$Substituting these values into the formula:$Z = \sqrt{4^2 + (9 - 6)^2}$$Z = \sqrt{16 + (3)^2}$$Z = \sqrt{16 + 9}$$Z = \sqrt{25}$$Z = 5 \ \Omega$Therefore,the impedance of the circuit is $5 \ \Omega$.

If a resistor of resistance $4 \Omega$,a capacitor of capacitive reactance $6 \Omega$ and an inductor of inductive reactance $9 \Omega$ are connected in series with an ac source,then the impedance of the circuit is (in $\Omega$)

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