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Chapter 32: Q. 55 (page 926)

The current lags the emf by 30° in a series RLC circuit with E0=10V and R=50Ω. What is the peak current through the circuit?

Short Answer

Expert verified

Peak current through RLC circuit is, 0.173A

Step by step solution

01

Given information

We are given that,

Peak voltage is, E0=10V

Resistance is, R=50Ω

Phase difference is,ϕ=30°

02

Simplify

Let inductive reactance is,XL

And capacitive reactance is,XC

We know that,

tanϕ=XL-XCRtan30°=XL-XC5013=XL-XC50XL-XC=503XL-XC2=5032XL-XC2=25003=833.3Ω

Therefore, Impedance is,

Z=R2+XL-XC2=502+833.3=2500+833.3=57.73Ω

Hence Peak current is,

localid="1650491817289" I0=E0Z=1057.73=0.173A

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Most popular questions from this chapter

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a 2.1mH inductor, and a 550nF capacitor. It is connected to a 50V rms oscillating voltage source with an adjustable frequency. An oscillating magnetic field is observed at a point 2.5mm from the center of one of the circuit wires, which is long and straight. What is the maximum magnetic field amplitude that can be generated?

A low-passRC filter with a crossover frequency of 1000Hzuses a 100Ωresistor. What is the value of the capacitor?

A low-pass filter consists of a 100mFcapacitor in series with a 159Ωresistor. The circuit is driven by an ACsource with a peak voltage of 5.00V.

a. What is the crossover frequency fc?

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