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Chapter 14: Problem 2075

The instantaneous voltage through a device of impedance \(20 \Omega\) is \(\varepsilon=80 \sin 100 \pi t\). The effective value of the current is, (a) \(3 \mathrm{~A}\) (b) \(2.828 \mathrm{~A}\) (c) \(1.732 \mathrm{~A}\) (d) \(4 \mathrm{~A}\)

Short Answer

Expert verified
The effective value of the current (also known as RMS current) for the given instantaneous voltage function is approximately \(2.828A\) (option b). This can be calculated by first finding the instantaneous current using Ohm's law, and then determining the RMS value using the sinusoidal function formula.

Step by step solution

01

Write down the given information

Instantaneous voltage, \(\varepsilon = 80\sin(100\pi t)\) Impedance, \(Z = 20\Omega\)
02

Find the instantaneous current using Ohm's law

Ohm's Law states that, \(V = IZ\), where V is voltage, I is current, and Z is the impedance. Instantaneous current is given by, \(i(t) = \frac{\varepsilon (t)}{Z}\) So, \(i(t)= \frac{80\sin(100\pi t)}{20}\) \(i(t)= 4\sin(100\pi t)\)
03

Calculate the effective (RMS) current

The formula for the effective (RMS) value of a sinusoidal function is given by, \(\text{RMS value} = \frac{(\text{maximum value})}{\sqrt{2}}\) We already have the instantaneous current, so we can determine the maximum current by observing the given function, \(i(t) = 4\sin(100\pi t)\) The maximum value of the current is 4 A (since the maximum value of \(\sin(100\pi t)\) is 1). Applying the formula RMS value, \(I_{rms} = \frac{4}{\sqrt{2}}\) \(I_{rms} = 2\sqrt{2}\) \(I_{rms} ≈ 2.828 A\) From the given options, the effective (RMS) value of the current is: \(I_{rms} = 2.828A\) (option b)

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Most popular questions from this chapter

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 \(14.14 \mathrm{amp}\)

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