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Elden [556K]
3 years ago
6

Consider a series RLC circuit where R = 957 Ω R=957 Ω and C = 8.25 μ F C=8.25 μF . However, the inductance L L of the inductor i

s unknown. To find its value, Anindya decides to perform some simple measurements. He applies an AC voltage that peaks at 84.0 V 84.0 V and observes that the resonance angular frequency occurs at 88500 rad/s 88500 rad/s .
Physics
1 answer:
natta225 [31]3 years ago
7 0

Answer:

Inductance of the circuit is equal to 6.30\times 10^{-4}H    

Explanation:

We have given resistance R = 957 ohm

Capacitance of the circuit C=8.25\mu F=8.25\times 10^{-6}F

Peak ac voltage is V_m=84volt

Angular frequency at resonance \omega =88500rad/sec

We have to find the value of inductance L

At resonance \omega L=\frac{1}{\omega C}

\omega ^2=\frac{1}{LC}

88500 ^2=\frac{1}{L\times 8.25\times 10^{-6}}

L=6.30\times 10^{-4}H

So inductance of the circuit is equal to 6.30\times 10^{-4}H

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The power delivered to the coil is 0.5 Watts

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The formula for calculating the power delivered to it is expressed according to the equation

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