An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance R are connected in seri

Question

An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance R are connected in series to an AC source of potential difference V volts. The potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of the current flowing through the LCR series circuit is $10\sqrt{2}$ A. The impedance of the circuit will be:

Accepted Answer

To find the impedance ($Z$), we first determine the total RMS voltage ($V$) of the source. In a series LCR circuit, this is given by the phasor sum: $V = \sqrt{V_R^2 + (V_L - V_C)^2}$. Substituting the given values, $V = \sqrt{40^2 + (40 - 10)^2} = \sqrt{1600 + 900} = 50$ V . The RMS current ($I$) is calculated from the peak current amplitude ($i_m$) using the relationship $I = \frac{i_m}{\sqrt{2}}$. Given $i_m = 10\sqrt{2}$ A, $I = \frac{10\sqrt{2}}{\sqrt{2}} = 10$ A . The impedance is then the ratio of the RMS voltage to the RMS current: $Z = \frac{V}{I} = \frac{50 	ext{ V}}{10 	ext{ A}} = 5 \ \Omega$ .

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