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Chapter 14: Problem 2082

A resistor \(30 \Omega\), inductor of reactance \(10 \Omega\) and the capacitor of reactance \(10 \Omega\) are connected in series to an ac voltage source \(\mathrm{e}=300 \sqrt{2} \sin (\omega \mathrm{t})\) The current in the circuit is (a) \(10 \sqrt{2 \mathrm{~A}}\) (b) \(10 \mathrm{~A}\) (c) \(30 \sqrt{11 \mathrm{~A}}\) (d) \((30 / \sqrt{11}) \mathrm{A}\)

Short Answer

Expert verified
The current in the circuit is (a) \(10\sqrt{2}\)A.

Step by step solution

01

Find the impedance of the circuit

: The impedance (Z) of a series RLC circuit is given by the formula: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] with R, X_L, and X_C representing the resistance, inductive reactance, and capacitive reactance respectively. Use the given values to calculate the impedance.
02

Plug in the given values and calculate the impedance

: Now, substitute the given values into the impedance formula: \[ Z = \sqrt{(30\Omega)^2 + (10\Omega - 10\Omega)^2} \] \[ Z = \sqrt{(30\Omega)^2 + (0\Omega)^2} \] \[ Z = 30\Omega \] The impedance of the circuit is 30Ω.
03

Use Ohm's Law to find the current

: Ohm's Law for an AC circuit is given by the formula: \[ I = \frac{V}{Z} \] where I is the current, V is the voltage, and Z is the impedance. Use the given voltage value, and the calculated impedance value to find the current.
04

Plug in the given values and calculate the current

: Now, substitute the given voltage value and the calculated impedance value into Ohm's Law: \[ I = \frac{300\sqrt{2}V}{30\Omega} \] \[ I = 10\sqrt{2}A \] So, the current in the circuit is \(10\sqrt{2}A\). Therefore, the correct answer is (a) \(10\sqrt{2}\)A.

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Most popular questions from this chapter

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