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

In an a.c. circuit the instantaneous current and emf are represented as $I = I_{0}, \sin [ω t - π / 6]$ and $E = E_{0} \sin [ω t + π / 3]$ respectively. The voltage leads the current by

A. $\frac{π}{2}$

B. $\frac{π}{4}$

C. $\frac{π}{3}$

D. $\frac{π}{6}$

Correct Option is (A)

Given $ϕ_{1} = \frac{π}{6}$ and $ϕ_{2} = \frac{π}{3}$

$∴ Δ ϕ = \frac{π}{3} - (\frac{- π}{6}) = \frac{π}{2}$

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

The coil of an a.c. generator has 100 turns, each of cross-sectional area $2 m^{2}$ . It is rotating at constant angular speed $30 rad / s$ , in a uniform magnetic field of $2 \times 10^{- 2} T$ . If the total resistance of the circuit is $600 Ω$ then maximum power dissipated in the circuit is

A. $6 W$

B. $9 W$

C. $12 W$

D. $24 W$

Correct Option is (C)

$\begin{aligned} N = 100, A = 2 m^{2}, ω = 30 rad / s \\ B = 2 \times 10^{- 2} T, R = 600 Ω \end{aligned}$

Maximum power dissipated in the circuit

$\begin{aligned} P_{max} = E_{r m s} \times I_{r m s} & = \frac{E_{0}}{\sqrt{2}} \times \frac{I_{0}}{\sqrt{2}} \\ = \frac{E_{0} I_{0}}{2} ..... (i) \end{aligned}$

But

I_{0} = \frac{E_{0}}{R}

..... (ii)

Putting (2) into (1) we get,

$I_{0} = \frac{E_{0}^{2}}{2 R}$

But $E_{0} = NAB ω$

$\begin{aligned} E_{0} & = 100 \times 2 \times 2 \times 10^{- 2} \times 30 \\ = 120 V \\ ∴ P_{max} = & \frac{120 \times 120}{2 \times 600} = 12 W \end{aligned}$

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

In an $AC$ circuit, the current is $i = 5 \sin (100 t - \frac{π}{2}) A$ and voltage is $e = 200 \sin (100 t)$ volt. Power consumption in the circuit is $(\cos 90^{\circ} = 0)$

A. $200 W$

B. $0 W$

C. $40 W$

D. $1000 W$

Correct Option is (B)

$\begin{aligned} P = I_{rms} V_{rms} \cos ϕ \\ ϕ = 90^{\circ} \\ ∴ & P = 0 (∵ \cos 90^{\circ} = 0) \end{aligned}$

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

The instantaneous value of an alternating current is given by $i = 50 \sin (100 π t)$ . It will achieve a value of $25 A$ after a time interval of $(\sin 30^{\circ} = 0.5)$

A. $\frac{1}{300} S$

B. $\frac{1}{100} S$

C. $\frac{1}{200} S$

D. $\frac{1}{600} S$

Correct Option is (B)

$\begin{aligned} i = 50 \sin 100 π t \\ ∴ & 25 = 50 \sin 100 π t \\ ∴ & \frac{25}{50} = \sin 100 π t or \frac{1}{2} = \sin 100 π t \\ ∴ & 100 π t = \frac{π}{6} ∴ t = \frac{1}{100} s \end{aligned}$

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

An alternating voltge is represented by $V = 80 \sin (100 π t) \cos (100 π t)$ volt. The peak voltage is

A. $20 V$

B. $40 V$

C. $30 V$

D. $50 V$

Correct Option is (B)

$\begin{aligned} \sin 2 θ = 2 \sin θ \cdot \cos θ \\ \begin{aligned} ∴ \sin θ \cos θ = \frac{\sin 2 θ}{2} \\ V = 80 \sin (100 π t) \cos (100 π t) \\ = \frac{80 \sin 200 π t}{2} = 40 \sin 200 π t \end{aligned} \end{aligned}$

Amplitude or the peak value of the voltage $= 40 V$

⇒AC Circuits