An L-C-R series circuit with L-0-120H- R-240- and C-7-30- F carries an rms current of 0-450A with a frequency of 400Hz- What is the impedance of the circuit-

R=240Ω, C=7.3μ F, L=0.12H,ω=2π×400=2512 Hz RMS current is i_r=0.45A #160;Impedance is given as Z=√(R^2+(ω L 1ω C)^2) . #160;So, Z=344Ω ...

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Standard XII

Physics

Time Constant

An L-C-R se...

Question

An $L - C - R$ series circuit with $L = 0.120 H, R = 240 Ω$ , and $C = 7.30 μ F$ carries an rms current of $0.450 A$ with a frequency of $400 H z$ . What is the impedance of the circuit?

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Solution

$R = 240 Ω, C = 7.3 μ F, L = 0.12 H, ω = 2 π \times 400 = 2512 H z$

RMS current is $i_{r} = 0.45 A$

Impedance is given as $Z = \sqrt{R^{2} + (ω L - \frac{1}{ω C})^{2}}$ .

So, $Z = 344 Ω$

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An L−C−R series circuit with L=0.120H,R=240Ω, and C=7.30μF carries an rms current of 0.450A with a frequency of 400Hz. What is the impedance of the circuit?

R=240Ω,C=7.3μF,L=0.12H,ω=2π×400=2512 HzRMS current is ir=0.45A Impedance is given as Z=√R2+(ωL−1ωC)2. So, Z=344Ω

An $L - C - R$ series circuit with $L = 0.120 H, R = 240 Ω$ , and $C = 7.30 μ F$ carries an rms current of $0.450 A$ with a frequency of $400 H z$ . What is the impedance of the circuit?

$R = 240 Ω, C = 7.3 μ F, L = 0.12 H, ω = 2 π \times 400 = 2512 H z$

RMS current is $i_{r} = 0.45 A$

Impedance is given as $Z = \sqrt{R^{2} + (ω L - \frac{1}{ω C})^{2}}$ .

So, $Z = 344 Ω$