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 as shown in figure. Potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of current flowing through LCR series circuit is 10√2 A. The impedance of the circuit is :

**Step 1 – Find the phasor values of the individual voltages**

The given potentials are peak (amplitude) values:

$V\_L^\{\backslash text\{peak\}\}\; =\; 40\backslash \; \backslash text\{V\},\backslash quad\; V\_C^\{\backslash text\{peak\}\}\; =\; 10\backslash \; \backslash text\{V\},\backslash quad\; V\_R^\{\backslash text\{peak\}\}\; =\; 40\backslash \; \backslash text\{V\}$

In a series LCR circuit the resistor voltage is in phase with the current, the inductor voltage leads the current by 90°, and the capacitor voltage lags the current by 90°. Hence the net reactive voltage is the difference between the inductor and capacitor voltages:

$V\_\{\backslash text\{reactive\}\}^\{\backslash text\{peak\}\}\; =\; V\_L^\{\backslash text\{peak\}\}\; -\; V\_C^\{\backslash text\{peak\}\}\; =\; 40\backslash \; \backslash text\{V\}\; -\; 10\backslash \; \backslash text\{V\}\; =\; 30\backslash \; \backslash text\{V\}$

**Step 2 – Resultant peak voltage of the source**

Using the right‑triangle of the phasor diagram, the magnitude of the total source voltage (peak) is

$V\_\{\backslash text\{total\}\}^\{\backslash text\{peak\}\}\; =\; \backslash sqrt\{(V\_R^\{\backslash text\{peak\}\})^2\; +\; (V\_\{\backslash text\{reactive\}\}^\{\backslash text\{peak\}\})^2\}\; =\; \backslash sqrt\{40^\{2\}\; +\; 30^\{2\}\}\; =\; \backslash sqrt\{2500\}\; =\; 50\backslash \; \backslash text\{V\}$

**Step 3 – Convert the current to its RMS value**

The problem states that the current amplitude (peak) is

$I^\{\backslash text\{peak\}\}\; =\; 10\backslash sqrt\{2\}\backslash \; \backslash text\{A\}$

Therefore the RMS current is

$I\_\{\backslash text\{rms\}\}\; =\; \backslash frac\{I^\{\backslash text\{peak\}\}\}\{\backslash sqrt\{2\}\}\; =\; \backslash frac\{10\backslash sqrt\{2\}\}\{\backslash sqrt\{2\}\}\; =\; 10\backslash \; \backslash text\{A\}$

**Step 4 – Calculate the magnitude of the impedance**

Impedance magnitude is the ratio of RMS voltage to RMS current. The RMS voltage corresponding to the peak value found in Step 2 is

$V\_\{\backslash text\{rms\}\}\; =\; \backslash frac\{V\_\{\backslash text\{total\}\}^\{\backslash text\{peak\}\}\}\{\backslash sqrt\{2\}\}\; =\; \backslash frac\{50\}\{\backslash sqrt\{2\}\}\; =\; 25\backslash sqrt\{2\}\backslash \; \backslash text\{V\}$

Thus

$Z\; =\; \backslash frac\{V\_\{\backslash text\{rms\}\}\}\{I\_\{\backslash text\{rms\}\}\}\; =\; \backslash frac\{25\backslash sqrt\{2\}\backslash \; \backslash text\{V\}\}\{10\backslash \; \backslash text\{A\}\}\; =\; \backslash frac\{25\backslash sqrt\{2\}\}\{10\}\backslash \; \backslash Omega\; =\; 5\backslash \; \backslash Omega$

**Result**: the impedance of the series LCR circuit is **5 Ω**.