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Class 12 NEET Physics Electromagnetic Induction and Alternating Currents 1

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Question : 11 of 100

Marks: +1, -0

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

R

40 \text{V}, 10

\text{V}

40 \text{V}

, respectively. The amplitude of current flowing through

angle C R

series circuit is

10 \sqrt{2} \text{A}

. The impedance of the circuit is 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.

$4 \sqrt{2} \Omega$
$\frac{5}{\sqrt{2}}\Omega$
$4 \Omega$
$5 \Omega$

Solution: 👈: Video Solution

V_{L}=40

volt

V_{R}=40

volt

V_{C}=10

volt

Now,

V_{R M S}=\sqrt{V_{R}^{2}+(V_{L}-V_{C})^{2}}

=\sqrt{(40)^{2}+(40-10)^{2}}=50 \text{V}

I_{R M S}=\frac{I_{0}}{\sqrt{2}}=\frac{10 \sqrt{2}}{\sqrt{2}}=10 \text{A}

\because V_{R M S}=I_{R M S} \times Z

\therefore Z=\frac{V_{R M S}}{I_{R M S}}=\frac{50}{10}=5 \Omega

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