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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 .

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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where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity.

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$W'=(m'+m)g=B$

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If we observe the field lines around a magnet, we observe that:

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7)

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8)

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See the last 7 answers in the attached document.

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> docx </span>

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>

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