Chapter 32: Q. 55 (page 926)

The current lags the emf by $30 °$ in a series RLC circuit with $E_{0} = 10 V$ and $R = 50 Ω$ . What is the peak current through the circuit?

Short Answer

Expert verified

Peak current through RLC circuit is, $0.173 A$

Step by step solution

Given information

We are given that,

Peak voltage is, $E_{0} = 10 V$

Resistance is, $R = 50 Ω$

Phase difference is, $ϕ = 30 °$

Simplify

Let inductive reactance is, $X_{L}$

And capacitive reactance is, $X_{C}$

We know that,

$\tan (ϕ) = \frac{X_{L} - X_{C}}{R} \tan (30 °) = \frac{X_{L} - X_{C}}{50} \frac{1}{\sqrt{3}} = \frac{X_{L} - X_{C}}{50} X_{L} - X_{C} = \frac{50}{\sqrt{3}} {(X_{L} - X_{C})}^{2} = {(\frac{50}{\sqrt{3}})}^{2} {(X_{L} - X_{C})}^{2} = \frac{2500}{3} = 833.3 Ω$

Therefore, Impedance is,

$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = \sqrt{50^{2} + 833.3} = \sqrt{2500 + 833.3} = 57.73 Ω$

Hence Peak current is,

localid="1650491817289" $I_{0} = \frac{E_{0}}{Z} = \frac{10}{57.73} = 0.173 A$

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The current lags the emf by $30 °$ in a series RLC circuit with $E_{0} = 10 V$ and $R = 50 Ω$ . What is the peak current through the circuit?

Short Answer

Step by step solution

Given information

Simplify

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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

Step by step solution

Given information

Simplify

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Scientific Method Physics

Electricity

Fluids

Waves Physics

Particle Model of Matter

Torque and Rotational Motion

Study anywhere. Anytime. Across all devices.

The current lags the emf by $30 °$ in a series RLC circuit with $E_{0} = 10 V$ and $R = 50 Ω$ . What is the peak current through the circuit?