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ipn [44]
3 years ago
7

The generator in a purely inductive AC circuit has an angular frequency of 363 rad/s. If the maximum voltage is 169 V and the in

ductance is 0.0937 H, what is the rms current in the circuit
Physics
1 answer:
irga5000 [103]3 years ago
5 0

Answer:

The rms current in the circuit is 3.513 A

Explanation:

Given;

angular frequency of the inductor, ω = 363 rad/s

maximum voltage of the inductive AC, V₀ = 169 V

Inductance of the inductor, L = 0.0937 H

Inductive reactance is given by;

X_L = 2\pi f L= \omega L

X_L = 363 *0.0937\\\\X_L = 34.0131 \ ohms

The rms voltage is given by;

V_{rms} = \frac{V_o}{\sqrt{2} } \\\\V_{rms} =\frac{169}{\sqrt{2} } \\\\V_{rms} = 119.5 \ V

The rms current in the circuit is given by;

I_{rms} = \frac{V_{rms}}{X_L} \\\\I_{rms} = \frac{119.5}{34.0131} \\\\I_{rms} = 3.513 \ A

Therefore, the rms current in the circuit is 3.513 A

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Answer:

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

Explanation:

u = Object distance

v = Image distance

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m = Magnification = 2.5

m=-\frac{v}{u}\\\Rightarrow 2.5=-\frac{v}{u}\\\Rightarrow v=-2.5 u

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\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{35}=\frac{1}{u}+\frac{1}{-2.5u}\\\Rightarrow \frac{1}{35}=\frac{3}{5u}\\\Rightarrow u=21\ cm

v=-2.5\times 21=-52.5\ cm

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

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