An alternating emf \(E = 440 \sin(100\pi t)\) is applied to a circuit containing an inductance of \(\frac{\sqrt{2}}{\pi}\) H. If an a.c. ammeter is connected in the circuit, its reading will be:
- 4.4 A
- 1.55 A
- 2.2 A
- 3.11 A
The Correct Option is C
Approach Solution - 1
To determine the reading on an AC ammeter connected to a circuit, we need to find the current \( I \) flowing through the circuit when the alternating EMF \( E \) is applied across an inductor.
The given EMF is:
\(E = 440 \sin(100\pi t)\)
The formula for current in an inductor when alternating EMF is applied is given by:
\(I = \frac{E_0}{Z}\)
where \( E_0 \) is the peak EMF and \( Z \) is the impedance of the inductor.
Since there is only an inductor in the circuit, the impedance \(Z = \omega L\), where:
- \(\omega\) is the angular frequency.
- \(L\) is the inductance.
From the given equation of EMF:
\(\omega = 100\pi\)
Given, the inductance \(L = \frac{\sqrt{2}}{\pi}\) H.
Therefore, the impedance:
\(Z = \omega L = 100\pi \times \frac{\sqrt{2}}{\pi} = 100\sqrt{2}\)
The peak EMF \( E_0 \) is given as 440 V.
Using the formula for current:
\(I = \frac{440}{100\sqrt{2}}\)
Calculate the current:
\(I = \frac{440}{100\sqrt{2}} = \frac{440}{141.42} \approx 3.11 \text{ A}\)
Since the ammeter measures the RMS value of the current, and for a sine wave:
\(I_{\text{rms}} = \frac{I_0}{\sqrt{2}}\)
Therefore, RMS current:
\(I_{\text{rms}} = \frac{3.11}{\sqrt{2}} = 2.2 \text{ A}\)
Thus, the reading on the AC ammeter will be 2.2 A, which matches the correct given option.
Approach Solution -2
Current \(I = \frac{V}{\omega L}\)
\(I = \frac{440}{100\pi \times \frac{\sqrt{2}}{\pi}}\)
\(I = \frac{44}{10\sqrt{2}}\)
\(⇒\) \(I_{\text{rms}} = \frac{I}{\sqrt{2}}\)
\(=\frac{44}{20}\)
\(=2.2 A\)
So, the correct option is (C): 2.2 A
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Concepts Used:
Alternating Current
An alternating current can be defined as a current that changes its magnitude and polarity at regular intervals of time. It can also be defined as an electrical current that repeatedly changes or reverses its direction opposite to that of Direct Current or DC which always flows in a single direction as shown below.
Alternating Current Production
Alternating current can be produced or generated by using devices that are known as alternators. However, alternating current can also be produced by different methods where many circuits are used. One of the most common or simple ways of generating AC is by using a basic single coil AC generator which consists of two-pole magnets and a single loop of wire having a rectangular shape.
Application of Alternating Current
AC is the form of current that are mostly used in different appliances. Some of the examples of alternating current include audio signal, radio signal, etc. An alternating current has a wide advantage over DC as AC is able to transmit power over large distances without great loss of energy.