16. ⇒ (MHT CET 2023 9th May Evening Shift )

The reactance of capacitor at $50 Hz$ is $5 Ω$ . If the frequency is increased to $100 Hz$ , the new reactance is

A. $5 Ω$

B. $10 Ω$

C. $2.5 Ω$

D. $125 Ω$

Correct Option is (C)

$\begin{aligned} X_{C} & = \frac{1}{2 π f c} \\ ∴ C & = \frac{1}{2 π f (5)} \dots . (∵ X_{C} = 5 Ω) \end{aligned}$

New reactance,

$X_{C}^{'} = \frac{1}{2 π f^{'} C} = \frac{1}{2 π (2 f) C} = \frac{X_{C}}{2} = \frac{1}{2} \times 5 = 2.5 Ω$

17. ⇒ (MHT CET 2023 9th May Morning Shift )

A capacitor, an inductor and an electric bulb are connected in series to an a.c. supply of variable frequency. As the frequency of the supply is increased gradually, then the electric bulb is found to

A. increase in brightness.

B. decrease in brightness.

C. increase, reach a maximum and then decrease in brightness.

D. show no change in brightness.

Correct Option is (C)

The given circuit acts like a series LCR circuit.

MHT CET 2023 9th May Morning Shift Physics - Alternating Current Question 24 English Explanation

$I = \frac{V}{Z} = \frac{V}{\sqrt{R_{2} + {(ω L - \frac{1}{ω C})}^{2}}}$

From the above formula, when $ω = 0$ , $I =$ maximum

When $ω$ is increased, I starts to decrease and when $ω = \frac{1}{\sqrt{LC}}$ the current will be zero. This cycle will repeat going forward. So brightness will increase, reach a maximum and then decrease.

18. ⇒ (MHT CET 2021 21th September Evening Shift )

The inductive reactance of a coil is R $Ω$ . If the inductance of a coil is doubled and frequency of a.c. supply is also doubled then the new inductive reactance will be

A. 2R

B. 8R

C. $\frac{R}{2}$

D. 4R

Correct Option is (D)

Inductive reactance = R = 2 $π$ fL

If L and f both are doubled, the inductive reactance will become 4R.

19. ⇒ (MHT CET 2021 21th September Morning Shift )

In the graphical representation of e.m.f. ' $e$ ' and current ' $i$ ' versus ' $ω t$ ' for an a.c. circuit, both emf and current reach zero, minimum and maximum value at the same time. The circuit element connected to the source will be

A. pure capacitor

B. combination of capacitor and inductor

C. pure resistor

D. pure inductor

Correct Option is (C)

Since the current and emf are in phase the circuit is purely resistive.

20. ⇒ (MHT CET 2021 21th September Morning Shift )

A series combination of resistor 'R' and capacitor 'C' is connected to an a.c. source of angular frequency ' $ω$ '. Keeping the voltage same, if the frequency is changed to $\frac{ω}{3}$ the current becomes half of the original current. Then the ratio of capacitive reactance and resistance at the former frequency is

A. $\sqrt{0.6}$

B. $\sqrt{6}$

C. $\sqrt{3}$

D. $\sqrt{2}$

Correct Option is (A)

Initial current $I = \frac{V}{z}$ , Final current $I^{'} = \frac{V}{z}$

$\begin{aligned} \frac{1}{2} = \frac{I^{'}}{I} = \frac{z}{z^{'}} ∴ \frac{z}{z^{'}} = \frac{1}{2} \\ ∴ \frac{z^{2}}{z^{' 2}} = \frac{1}{4} ∴ \frac{R^{2} + X_{c}^{2}}{R^{2} + X_{c}^{' 2}} = \frac{1}{4} \\ ∴ 4 R^{2} + 4 X_{c}^{2} = R^{2} + X_{c}^{' 2} \\ ∴ 3 R^{2} = X_{c}^{' 2} - 4 X_{c}^{2} ......(1) \end{aligned}$

$X_{c} = \frac{1}{ω C}, X^{'} = \frac{3}{ω C} ∴ X^{'} = 3 X_{c}$

Putting this value of $X^{'}$ in (1) we get

$3 R^{2} = 9 X_{c}^{2} - 4 X_{c}^{2} = 5 X_{c}^{2}$

$\begin{aligned} ∴ \frac{X_{c}^{2}}{R^{2}} = \frac{3}{5} = 0.6 \\ ∴ \frac{X_{c}}{R} = \sqrt{0.6} \end{aligned}$

⇒AC Circuits