Question

In a series LCR circuit, V_L = 40 V, V_C = 10 V, V_R = 40 V, and current amplitude is 10\[\sqrt 2\]​ A. Find the impedance.

Options

• 5 Ω

• 4\[\sqrt 2\] Ω

• \[\frac {5}{\sqrt 2}\]Ω

• 4 Ω

Answer 1

5 Ω

Explanation:

Given, Inductance of the inductor = L
Capacitance of the capacitor = C
Resistance of the resistor = R
Potential difference = V

I₀ = 10\[\sqrt 2\]A

\[V_{\mathrm{rms}}=\sqrt{\mathrm{v}_{R}^{2}+\left(\mathrm{V}_{L}-\mathrm{V}_{C}\right)^{2}}\]

Now, substituting the values,

\[V_{\mathrm{rms}}=\sqrt{40^{2}+\left(40-10\right)^{2}}\] = \[\sqrt{40^2+30^2}\] = \[\sqrt{1600+900}\] = \[\sqrt {2500}\] = 50 V

\[I_{rms}=\frac{I_0}{\sqrt{2}}=\frac{10\sqrt{2}}{\sqrt{2}}\] = 10 A

\[V_{\mathrm{rms}}=I_{\mathrm{rms}}\times Z\]

Z = \[\frac {V_{rms}}{I_{rms}}\] = \[\frac {50}{10}\] = 5 Ω

Hence, the required impedance of the circuit = 5 Ω.