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Alternating Current — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current4 Marks
Question
Let a source of alternating e.m.f. $\text{E} = \text{E}_\circ\sin\omega\text{t}$ be connected to a circuit containing a pure inductance L. If I is the value of instantaneous current in the circuit, then $\text{I}=\text{I}_\circ\sin\Big(\omega\text{t}-\frac{\pi}{2}\Big).$ The inductive reactance limits the current in a purely inductive circuit and is given by $\text{X}_\text{L}= \text{W}_\text{L}.$
  1. A 100 hertz a.c. is flowing in a 14mH coil. The reactance is:
  1. $15\Omega$
  2. $7.7\Omega$
  3. $8.8\Omega$
  4. $10\Omega$
  1. In a pure inductive circuit, resistance to the flow of current is offered by:
  1. Resistor
  2. Inductor
  3. Capacitor
  4. Resistor and inductor
  1. In a inductive circuit, by what value of phase angle does alternating current lags behind e.m.f.?
  1. 45º
  2. 90º
  3. 120º
  4. 75º
  1. How much inductance should be connected to 200V, 50Hz a.c. supply so that a maximum current of 0.9A flows through it?
  1. 5H
  2. 1H
  3. 10H
  4. 4.5H
  1. The maximum value of current when inductance of 2H is connected to 150V, 50Hz supply is:
  1. 0.337A
  2. 0.721A
  3. 1.521A
  4. 2.522A
✓

Answer

  1. (c) $8.8\Omega$
Explanation:
Inductive reactance,
$\text{X}_\text{L}= \text{W}_\text{L}=2\pi\mu\text{L}$
$=2\pi\times100\times14\times10^{-3}$
$\text{X}_\text{L}=8.8\Omega$
  1. (b) Inductor
  2. (b) 90º
Explanation:
In an inductor vol tag: leads the current by $\frac{\pi}{2}$ or current lags the voltage by $\frac{\pi}{2}.$
  1. (b) 1H
Explanation:
The current in the inductor coil is given by,
$\text{I}_0=\frac{\text{E}_0}{\text{X}_\text{L}}=\frac{\sqrt{2}\text{E}_\text{V}}{2\pi\mu\text{L}}$
$\text{L}=\frac{\sqrt{2}\text{E}_\text{V}}{2\pi\mu\text{I}_0}=\frac{1.414\times200}{2\times3.14\times50\times0.9}=1\text{H}$
  1. (a) 0.337A
Explanation:
Inductive reactance,
$\text{X}_\text{L}= \text{W}_\text{L}=2\pi\mu\text{L}$
$=2\pi\times3.14\times50\times2=628\Omega$
$\text{I}_0=\frac{\text{E}_0}{\text{X}_\text{L}}$
$\Rightarrow\text{I}_0=\frac{\sqrt{2}\times\text{E}_\text{V}}{\text{X}_\text{L}}=\frac{\sqrt{2}\times150}{628}=0.337\text{A}$

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