1. ⇒ (MHT CET 2023 13th May Morning Shift )

With increase in frequency of a.c. supply, the impedance of an L-C-R series circuit

A. remains constant.

B. increases.

C. decreases.

D. decreases at first, becomes minimum and then increases.

Correct Option is (D)

We know

$X_{L} = L ω and X_{C} = \frac{1}{C ω}$

$\Rightarrow$ When the frequency increases, $X_{L}$ increases and $X_{C}$ decreases.

$∴$ The impedance of an LCR series circuit decreases at first, becomes minimum and then increases.

2. ⇒ (MHT CET 2023 12th May Evening Shift )

When an inductor ' $L$ ' and a resistor ' $R$ ' in series are connected across a $15 V, 50 Hz$ a.c. supply, a current of $0.3 A$ flows in the circuit. The current differs in phase from applied voltage by ${(\frac{π}{3})}^{c}$ . The value of ' $R$ ' is $(\sin \frac{π}{6} = \cos \frac{π}{3} = \frac{1}{2}, \sin \frac{π}{3} = \cos \frac{π}{6} = \frac{\sqrt{3}}{2})$

A. $10 Ω$

B. $15 Ω$

C. $20 Ω$

D. $25 Ω$

Correct Option is (D)

$\begin{array}{ll} Given: E_{v} = 15 V, f = 50 Hz, I = 0.3 A, \\ ϕ = \frac{π}{3} rad \\ Impedance Z = \frac{E_{v}}{I} = \frac{15}{0.2} = 50 Ω \\ \tan ϕ = \frac{X_{L}}{R} \\ \tan \frac{π}{3} = \frac{X_{L}}{R} \\ \sqrt{3} = \frac{X_{L}}{R} \\ ∴ & X_{L} = \sqrt{3} R \\ Impedance Z = \sqrt{R^{2} + X_{L}^{2}} \\ Z = \sqrt{R^{2} + (\sqrt{3} R)^{2}} \\ Z = \sqrt{4 R^{2}} \\ ∴ & 2 R = Z \\ ∴ & R = \frac{Z}{2} = \frac{50}{2} = 25 Ω \end{array}$

3. ⇒ (MHT CET 2023 12th May Evening Shift )

An a.c. source of $15 V, 50 Hz$ is connected across an inductor (L) and resistance (R) in series R.M.S. current of $0.5 A$ flows in the circuit. The phase difference between applied voltage and current is $(\frac{π}{3})$ radian. The value of resistance $(R)$ is $(\tan 60^{\circ} = \sqrt{3})$

A. $10 Ω$

B. $12 Ω$

C. $15 Ω$

D. $20 Ω$

Correct Option is (C)

Given data: $E = 15 V, f = 50 Hz, I = 0.5 A, ϕ = \frac{π}{3} rad$

Impedance is given as $Z = \frac{E}{I} = \frac{15}{0.5} = 30 Ω$

$\begin{matrix} \tan ϕ = \frac{X_{L}}{R} \\ \tan \frac{π}{3} = \frac{X_{L}}{R} \\ \sqrt{3} = \frac{X_{L}}{R} \\ ∴ X_{L} = \sqrt{3} R \end{matrix}$

The formula for impedance is

$\begin{aligned} Z & = \sqrt{R^{2} + X_{L}^{2}} \\ Z & = \sqrt{R^{2} + (\sqrt{3} R)^{2}} \\ Z & = \sqrt{4 R^{2}} \\ ∴ 2 R & = Z \\ ∴ R & = \frac{Z}{2} = \frac{30}{2} = 15 Ω \end{aligned}$

4. ⇒ (MHT CET 2023 12th May Morning Shift )

In the given circuit, r.m.s. value of current through the resistor $R$ is

MHT CET 2023 12th May Morning Shift Physics - Alternating Current Question 6 English

A. $2 A$

B. $0.5 A$

C. $20 A$

D. $2 \sqrt{2} A$

Correct Option is (A)

$\begin{aligned} Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} \\ Z = \sqrt{100^{2} + 100^{2}} \\ Z = 100 \sqrt{2} Ω \\ i_{rms} = \frac{V_{rms}}{Z} \\ ∴ i_{rms} = \frac{200 \sqrt{2}}{100 \sqrt{2}} \\ i_{rms} = 2 A \end{aligned}$

5. ⇒ (MHT CET 2023 12th May Morning Shift )

An inductor of $0.5 mH$ , a capacitor of $20 μ F$ and a resistance of $20 Ω$ are connected in series with a $220 V$ a.c. source. If the current is in phase with the e.m.f. the maximum current in the circuit is $\sqrt{x} A$ . The value of ' $x$ ' is

A. 44

B. 82

C. 146

D. 242

Correct Option is (D)

When current is in phase with voltage, we have

$\begin{aligned} Z = R = 20 Ω \\ e_{0} = \sqrt{2} e_{rms} = 220 \sqrt{2} V \\ i_{0} = \frac{e_{0}}{Z} = \frac{220 \sqrt{2}}{20} = 11 \sqrt{2} A \\ i_{0} = \sqrt{242} A \end{aligned}$

⇒AC Circuits