QBManager

AC (NCERT)

274678 In an a.c. circuit $V$ and $I$ are given by
$\begin{array}{r} V = 100 sin (100 t) volts \\ I = 100 sin (100 t + π / 3) mA \end{array}$
the power dissipated in the circuit is

1

10^{4}

watt

2 10 watt

3 2.5watt

4 5.0 watt

Explanation:

(c) $P = V_{r . m . s} \times I_{r . m . s} \times cos ϕ = \frac{1}{2} V_{0} I_{0} cos ϕ$
$= \frac{1}{2} \times 100 \times (100 \times 10^{- 3}) cos π / 3 = 2.5 W$

NCERT Page-252/ N-191

AC (NCERT)

274679 The instantaneous values of alternating current and voltages in a circuit are given as
$\begin{array}{r} i = \frac{1}{\sqrt{2}} sin (100 π t) A \\ e = \frac{1}{\sqrt{2}} sin (100 π t + π / 3) Volt \end{array}$
The average power in Watt consumed in the circuit is

1

\frac{1}{4}

2

\frac{\sqrt{3}}{4}

3

\frac{1}{2}

4

\frac{1}{8}

Explanation:

(d) The average power in the circuit where $cos ϕ =$ power factory $P = V_{rms} \times I_{rms} cos ϕ$
$ϕ = π / 3 =$ phase difference $= \frac{180}{3} = 60$
$V_{rms} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} = \frac{1}{2} volt; I_{rms} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} = (\frac{1}{2}) A$
$⟨ P ⟩ = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} W$

NCERT Page-252 / N-191

AC (NCERT)

274680 Two coils A and B are connected in series across a $240 V$ , $50 Hz$ supply. The resistance of $A$ is $5 Ω$ and the inductance of $B$ is $0.02 H$ . The power consumed is $3 kW$ and the power factor is 0.75 . The impedance of the circuit is

1

0.144 Ω

2

1.44 Ω

3

14.4 Ω

4

144 Ω

Explanation:

(c) $P = \frac{E_{v}^{2} cos ϕ}{Z}$

NCERT Page-252 / N-191

AC (NCERT)

274681 In an electrical circuit $R, L, C$ and an a.c. voltage source are all connected in series. When $L$ is removed from the circuit, the phase difference between the voltage the current in the circuit is $π / 3$ . If instead, $C$ is removed from the circuit, the phase difference is again $π / 3$ . The power factor of the circuit is

1

1 / 2

2

1 / \sqrt{2}

3 1

4

\sqrt{3} / 2

Explanation:

(c)
$P = 3000 = \frac{{(240)}^{2} (0.75)}{Z} d \Rightarrow Z = 14.4 Ω$

NCERT Page-252 / N-191

AC (NCERT)

274678 In an a.c. circuit $V$ and $I$ are given by
$\begin{array}{r} V = 100 sin (100 t) volts \\ I = 100 sin (100 t + π / 3) mA \end{array}$
the power dissipated in the circuit is

1

10^{4}

watt

2 10 watt

3 2.5watt

4 5.0 watt

Explanation:

(c) $P = V_{r . m . s} \times I_{r . m . s} \times cos ϕ = \frac{1}{2} V_{0} I_{0} cos ϕ$
$= \frac{1}{2} \times 100 \times (100 \times 10^{- 3}) cos π / 3 = 2.5 W$

NCERT Page-252/ N-191

AC (NCERT)

274679 The instantaneous values of alternating current and voltages in a circuit are given as
$\begin{array}{r} i = \frac{1}{\sqrt{2}} sin (100 π t) A \\ e = \frac{1}{\sqrt{2}} sin (100 π t + π / 3) Volt \end{array}$
The average power in Watt consumed in the circuit is

1

\frac{1}{4}

2

\frac{\sqrt{3}}{4}

3

\frac{1}{2}

4

\frac{1}{8}

Explanation:

(d) The average power in the circuit where $cos ϕ =$ power factory $P = V_{rms} \times I_{rms} cos ϕ$
$ϕ = π / 3 =$ phase difference $= \frac{180}{3} = 60$
$V_{rms} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} = \frac{1}{2} volt; I_{rms} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} = (\frac{1}{2}) A$
$⟨ P ⟩ = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} W$

NCERT Page-252 / N-191

AC (NCERT)

274680 Two coils A and B are connected in series across a $240 V$ , $50 Hz$ supply. The resistance of $A$ is $5 Ω$ and the inductance of $B$ is $0.02 H$ . The power consumed is $3 kW$ and the power factor is 0.75 . The impedance of the circuit is

1

0.144 Ω

2

1.44 Ω

3

14.4 Ω

4

144 Ω

Explanation:

(c) $P = \frac{E_{v}^{2} cos ϕ}{Z}$

NCERT Page-252 / N-191

AC (NCERT)

274681 In an electrical circuit $R, L, C$ and an a.c. voltage source are all connected in series. When $L$ is removed from the circuit, the phase difference between the voltage the current in the circuit is $π / 3$ . If instead, $C$ is removed from the circuit, the phase difference is again $π / 3$ . The power factor of the circuit is

1

1 / 2

2

1 / \sqrt{2}

3 1

4

\sqrt{3} / 2

Explanation:

(c)
$P = 3000 = \frac{{(240)}^{2} (0.75)}{Z} d \Rightarrow Z = 14.4 Ω$

NCERT Page-252 / N-191

AC (NCERT)

274678 In an a.c. circuit $V$ and $I$ are given by
$\begin{array}{r} V = 100 sin (100 t) volts \\ I = 100 sin (100 t + π / 3) mA \end{array}$
the power dissipated in the circuit is

1

10^{4}

watt

2 10 watt

3 2.5watt

4 5.0 watt

Explanation:

(c) $P = V_{r . m . s} \times I_{r . m . s} \times cos ϕ = \frac{1}{2} V_{0} I_{0} cos ϕ$
$= \frac{1}{2} \times 100 \times (100 \times 10^{- 3}) cos π / 3 = 2.5 W$

NCERT Page-252/ N-191

AC (NCERT)

274679 The instantaneous values of alternating current and voltages in a circuit are given as
$\begin{array}{r} i = \frac{1}{\sqrt{2}} sin (100 π t) A \\ e = \frac{1}{\sqrt{2}} sin (100 π t + π / 3) Volt \end{array}$
The average power in Watt consumed in the circuit is

1

\frac{1}{4}

2

\frac{\sqrt{3}}{4}

3

\frac{1}{2}

4

\frac{1}{8}

Explanation:

(d) The average power in the circuit where $cos ϕ =$ power factory $P = V_{rms} \times I_{rms} cos ϕ$
$ϕ = π / 3 =$ phase difference $= \frac{180}{3} = 60$
$V_{rms} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} = \frac{1}{2} volt; I_{rms} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} = (\frac{1}{2}) A$
$⟨ P ⟩ = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} W$

NCERT Page-252 / N-191

AC (NCERT)

274680 Two coils A and B are connected in series across a $240 V$ , $50 Hz$ supply. The resistance of $A$ is $5 Ω$ and the inductance of $B$ is $0.02 H$ . The power consumed is $3 kW$ and the power factor is 0.75 . The impedance of the circuit is

1

0.144 Ω

2

1.44 Ω

3

14.4 Ω

4

144 Ω

Explanation:

(c) $P = \frac{E_{v}^{2} cos ϕ}{Z}$

NCERT Page-252 / N-191

AC (NCERT)

274681 In an electrical circuit $R, L, C$ and an a.c. voltage source are all connected in series. When $L$ is removed from the circuit, the phase difference between the voltage the current in the circuit is $π / 3$ . If instead, $C$ is removed from the circuit, the phase difference is again $π / 3$ . The power factor of the circuit is

1

1 / 2

2

1 / \sqrt{2}

3 1

4

\sqrt{3} / 2

Explanation:

(c)
$P = 3000 = \frac{{(240)}^{2} (0.75)}{Z} d \Rightarrow Z = 14.4 Ω$

NCERT Page-252 / N-191

AC (NCERT)

274678 In an a.c. circuit $V$ and $I$ are given by
$\begin{array}{r} V = 100 sin (100 t) volts \\ I = 100 sin (100 t + π / 3) mA \end{array}$
the power dissipated in the circuit is

1

10^{4}

watt

2 10 watt

3 2.5watt

4 5.0 watt

Explanation:

(c) $P = V_{r . m . s} \times I_{r . m . s} \times cos ϕ = \frac{1}{2} V_{0} I_{0} cos ϕ$
$= \frac{1}{2} \times 100 \times (100 \times 10^{- 3}) cos π / 3 = 2.5 W$

NCERT Page-252/ N-191

AC (NCERT)

274679 The instantaneous values of alternating current and voltages in a circuit are given as
$\begin{array}{r} i = \frac{1}{\sqrt{2}} sin (100 π t) A \\ e = \frac{1}{\sqrt{2}} sin (100 π t + π / 3) Volt \end{array}$
The average power in Watt consumed in the circuit is

1

\frac{1}{4}

2

\frac{\sqrt{3}}{4}

3

\frac{1}{2}

4

\frac{1}{8}

Explanation:

(d) The average power in the circuit where $cos ϕ =$ power factory $P = V_{rms} \times I_{rms} cos ϕ$
$ϕ = π / 3 =$ phase difference $= \frac{180}{3} = 60$
$V_{rms} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} = \frac{1}{2} volt; I_{rms} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} = (\frac{1}{2}) A$
$⟨ P ⟩ = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} W$

NCERT Page-252 / N-191

AC (NCERT)

274680 Two coils A and B are connected in series across a $240 V$ , $50 Hz$ supply. The resistance of $A$ is $5 Ω$ and the inductance of $B$ is $0.02 H$ . The power consumed is $3 kW$ and the power factor is 0.75 . The impedance of the circuit is

1

0.144 Ω

2

1.44 Ω

3

14.4 Ω

4

144 Ω

Explanation:

(c) $P = \frac{E_{v}^{2} cos ϕ}{Z}$

NCERT Page-252 / N-191

AC (NCERT)

274681 In an electrical circuit $R, L, C$ and an a.c. voltage source are all connected in series. When $L$ is removed from the circuit, the phase difference between the voltage the current in the circuit is $π / 3$ . If instead, $C$ is removed from the circuit, the phase difference is again $π / 3$ . The power factor of the circuit is

1

1 / 2

2

1 / \sqrt{2}

3 1

4

\sqrt{3} / 2

Explanation:

(c)
$P = 3000 = \frac{{(240)}^{2} (0.75)}{Z} d \Rightarrow Z = 14.4 Ω$

NCERT Page-252 / N-191

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