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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

An inductor of inductance \(\mathrm{L}\) and resistor of resistance \(\mathrm{R}\) are joined in series and connected by a source of frequency 0 power dissipated in the circuit is, (a) $\left[\left\\{\mathrm{R}^{2}+\mathrm{c}^{2} \mathrm{~L}^{2}\right\\} / \mathrm{V}\right]$ (b) $\left[\left\\{\mathrm{V}^{2} \mathrm{R}\right\\} /\left\\{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}\right\\}\right]$ (c) $\left[\mathrm{V} /\left\\{\mathrm{R}^{2}+\omega^{2} \mathrm{~L}^{2}\right\\}\right]$ (d) $\left[\sqrt{ \left.\left\\{R^{2}+\omega^{2} L^{2}\right\\} / V^{2}\right]}\right.$

The rms value of an ac current of \(50 \mathrm{~Hz}\) is 10 amp. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be, (a) \(2 \times 10^{-2}\) sec and \(14.14 \mathrm{amp}\) (b) \(1 \times 10^{-2}\) sec and \(7.07 \mathrm{amp}\) (c) \(5 \times 10^{-3}\) sec and \(7.07\) amp (d) \(5 \times 10^{-3} \mathrm{sec}\) and \(14.14 \mathrm{amp}\)

A 20 volts ac is applied to a circuit consisting of a resistance and a coil with negligible resistance. If the voltage across the resistance is $12 \mathrm{~V}$, the voltage across the coil is, (a) 16 volts (b) 10 volts (c) 8 volts (d) 6 volts

A rectangular loop with a sliding rod of length \(2 \mathrm{~m} \&\) resistance \(2 \Omega\). It moves in a uniform magnetic field of \(3 \mathrm{~T}\) perpendicular to plane of loop. The external force required to keep the rod moving with constant velocity of \(2 \mathrm{~ms}^{-1}\) is

Two similar circular loops carry equal currents in the same direction. On moving the coils further apart, the electric current will (a) Remain unchanged (b) Increasing in both (c) Increasing in one decreasing in other (d) Decreasing in both
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