SATHEE: Electromagnetic Induction and Alternating Current 6 Question 1

Electromagnetic Induction and Alternating Current 6 Question 1

1. A circuit connected to an $A C$ source of emf $e = e_{0} \sin (100 t)$ with $t$ in seconds, gives a phase difference of $\frac{π}{4}$ between the emf $e$ and current $i$ . Which of the following circuits will exhibit this?

(Main 2019, 8 Apr II)

(a) $R C$ circuit with $R = 1 k Ω$ and $C = 1 μ F$

(b) $R L$ circuit with $R = 1 k Ω$ and $L = 1 m H$

(d) $R L$ circuit with $R = 1 k Ω$ and $L = 10 m H$

Show Answer

Answer:

Correct Answer: 1. (c)

Solution:

Given, phase difference, $φ = \frac{π}{4}$

As we know, for $R - L$ or $R - C$ circuit,

$\begin{aligned} \Rightarrow \tan \frac{π}{4} & = \frac{X_{C} or X_{L}}{R} \\ 1 & = \frac{X_{C} or X_{L}}{R} \Rightarrow R = X_{C} or X_{L} \end{aligned}$

Also, given $e = e_{0} \sin (100 t)$

Comparing the above equation with general equation of emf, i.e. $e = e_{0} \sin ω t$ , we get

$ω = 100 r a d / s = 10^{2} r a d / s$

Now, checking option wise,

For $R - C$ circuit, with

$\begin{aligned} R & = 1 k Ω = 10^{3} Ω and C = 1 μ F = 10^{- 6} F \\ So, X_{C} & = \frac{1}{ω C} = \frac{1}{10^{2} \times 10^{- 6}} = 10^{4} Ω \Rightarrow R \neq X_{C} \end{aligned}$

For $R$ - $L$ circuit, with

$\begin{matrix} R = 1 k Ω = 10^{3} Ω \\ and L = 1 m H = 10^{- 3} H \\ So, X_{L} = ω L = 10^{2} \times 10^{- 3} = 10^{- 1} Ω \Rightarrow R \neq X_{L} \end{matrix}$

For $R$ - $C$ circuit, with

$\begin{array}{ll} R = 1 k Ω = 10^{3} Ω \\ and & C = 10 μ F = 10 \times 10^{- 6} F = 10^{- 5} F \\ So, & X_{C} = \frac{1}{10^{2} \times 10^{- 5}} = 10^{3} Ω \Rightarrow R = C \end{array}$

For $R$ - $L$ circuit, with

$\begin{aligned} R & = 1 k Ω = 10^{3} Ω \\ L & = 10 m H = 10 \times 10^{- 3} H = 10^{- 2} H \\ X_{L} & = 10^{2} \times 10^{- 2} = 1 Ω \Rightarrow R \neq X_{L} \end{aligned}$

and

Alternate Solution

Since, $\tan \frac{π}{4} = 1 = \frac{X_{C} or X_{L}}{R}$

$∴$ For $R - C$ circuit, we have

$1 = \frac{1}{C ω R} or ω = \frac{1}{C R}$

Similarly, for $R - L$ circuit, we have

$1 = \frac{ω L}{R} \Rightarrow ω = \frac{R}{L}$

It is given in the question that, $ω = 100 r a d / s$

Thus, again by substituting the given values of $R, C$ or $L$ option wise in the respective Eqs. (i) and (ii), we get that only for option (c),

$ω = \frac{1}{C R} = \frac{1}{10 \times 10^{- 6} \times 10^{3}} or ω = 100 r a d / s$

पिछला

अगला

Electromagnetic Induction and Alternating Current 6 Question 1

1. A circuit connected to an $A C$ source of emf $e = e_{0} \sin (100 t)$ with $t$ in seconds, gives a phase difference of $\frac{π}{4}$ between the emf $e$ and current $i$ . Which of the following circuits will exhibit this?

Answer:

Alternate Solution

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Electromagnetic Induction and Alternating Current 6 Question 1

1. A circuit connected to an AC source of emf e=e0sin⁡(100t) with t in seconds, gives a phase difference of π4 between the emf e and current i. Which of the following circuits will exhibit this?

Answer:

Alternate Solution

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Menu

1. A circuit connected to an $A C$ source of emf $e = e_{0} \sin (100 t)$ with $t$ in seconds, gives a phase difference of $\frac{π}{4}$ between the emf $e$ and current $i$ . Which of the following circuits will exhibit this?