SATHEE: alternating-currents Question 22

alternating-currents Question 22

Question: Q. 8. A $200 mH$ (pure) inductor and a $5 μ F$ (pure) capacitor are connected one by one, across a sinusoidal $a c$ voltage source of $V = [70.7 \sin (1000$ $t)]$ voltage. Obtain the expression for the current in each case.

A [Foreign, 2016]

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Solution:

Ans. For the applied voltage

$\begin{aligned} V & = 70.7 \sin (1000 t), we have V_{0} & = 70.7 volts \end{aligned}$

For the inductor

$\begin{aligned} i_{o} & = \frac{V_{0}}{ω L} = \frac{70.7}{1000 \times 200 \times 10^{- 3}} A & = 35.35 \times 10^{- 2} A & = 0.3535 A \end{aligned}$

$∴$ Expression for current is

$\begin{matrix} (Q.) & i = (0.3535) \sin (1000 t - \frac{π}{2}) \end{matrix}$

For the capacitor

$\begin{aligned} i_{0} & = \frac{V_{0}}{(\frac{1}{ω C})} = V_{0} \cdot ω C & = 70.7 \times 1000 \times 5 \times 10^{- 6} A & = 353.5 \times 10^{- 3} A = 0.3535 A \end{aligned}$

$∴$ Expression for current is

$I = 0.3535 \sin (1000 t + \frac{π}{2})$

[CBSE Marking Scheme, 2016]

[AI Q. 9. A circuit containing an $80 mH$ inductor and a $250 μ F$ capacitor in series connected to a $240 V$ , $100 rad / s$ supply. The resistance of the circuit is negligible.

(i) Obtain $r m s$ value of current.

(ii) What is the total average power consumed by the circuit?

A [Delhi I, II, III 2015]

Ans. (i)

$\begin{aligned} X_{L} & = ω L = 100 \times 80 \times 10^{- 3} = X_{C} & = \frac{1}{ω C} = \frac{1}{100 \times 250 \times 10^{- 6}} & = 40 Ω \end{aligned}$

Total Impedance $(Z) = X_{C} - X_{L}$

$= 32 Ω$

$I_{r m s} = \frac{240}{32} A \Leftrightarrow 7.5 A$

Alternating Currents

alternating-currents Question 22

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Alternating Currents

alternating-currents Question 22

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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