Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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 $R$ are 40 V, 10 V and 40 V, respectively. The amplitude of current flowing through $LCR$ series circuit is $10\sqrt{2}$ A. The impedance of the circuit is

$4\sqrt{2} \Omega$

$\frac{5}{\sqrt{2}} \Omega$

$4 \Omega$

$5 \Omega$

Step-by-Step Solution

The total voltage $V = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{40^2 + (40 - 10)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50$ V. The impedance $Z = \frac{V}{I_{rms}}$ . Since $I_{peak} = 10\sqrt{2}$ A, $I_{rms} = 10$ A. Thus, $Z = \frac{50}{10} = 5 \Omega$ .

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