155331
An inductor and a resistor are connected in series to an ac source of variable frequency. the power factor of the circuit is $\frac{\sqrt{3}}{2}$. If the frequency of the ac is increased by $200 \%$, the power factor of the circuit is
1 0.8
2 0.9
3 0.7
4 0.5
Explanation:
D According to question, Given, power factor $(\cos \phi)=\frac{\sqrt{3}}{2}$ $\mathrm{f}=50 \mathrm{~Hz}$ $\mathrm{f}^{\prime}=\mathrm{f}+200 \% \text { of } \mathrm{f}$ $\mathrm{f}^{\prime}=\mathrm{f}+\frac{200}{100} \times \mathrm{f}$ $\mathrm{f}^{\prime}=3 \mathrm{f}$ For R, L circuit $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+X_{L}^{2}}}$ $\cos ^{2} \phi=\frac{R^{2}}{R^{2}+X_{L}^{2}}$ $\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{R^{2}}{R^{2}+X_{L}^{2}}$ $\frac{3}{4}=\frac{R^{2}}{R^{2}+X_{L}^{2}}$ $3 R^{2}+3 X_{L}^{2}=4 R^{2}$ $R^{2}=3 \times X_{L}^{2}$ $X_{L}=\frac{R}{\sqrt{3}}$ $\because \quad \mathrm{X}_{\mathrm{L}} \propto \mathrm{f}$ $\therefore \quad$ After change in frequency, $\mathrm{X}_{\mathrm{L}}^{\prime}=3 \mathrm{~L}$ $\mathrm{X}_{\mathrm{L}}^{\prime}=3 \times \frac{\mathrm{R}}{\sqrt{3}}$ $\therefore \quad \cos \phi^{\prime} =\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{\prime 2}}}$ $\cos \phi^{\prime} =\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+(\sqrt{3} \mathrm{R})^{2}}}$ When the frequency of the applied ac is $50 \mathrm{~Hz}$, $\cos \phi^{\prime} =\frac{\mathrm{R}}{\sqrt{4} \mathrm{R}^{2}}$ $\cos \phi^{\prime} =\frac{\mathrm{R}}{2 \mathrm{R}}$ $\cos \phi^{\prime} =0.5$
AP EAMCET-24.04.2017
Alternating Current
155332
What is power dissipation in an A.C. circuit in which voltage and current are given by $V=300 \sin (\omega t+\pi / 2), I=5 \sin \omega t$ ?
155333
In an AC circuit, current is $3 \mathrm{~A}$ and voltage 210 $\mathrm{V}$ and power is $63 \mathrm{~W}$. The power factor is
1 0.11
2 0.09
3 0.08
4 0.10
Explanation:
D Given, $\mathrm{I}=3 \mathrm{~A}$ $\mathrm{~V}=210 \mathrm{~V}$ $\mathrm{P}=63 \mathrm{~W}$ $\because$ Power drawn by the circuit is - $\mathrm{P}=\mathrm{IV} \cos \phi$ Power factor, $(\cos \phi)=\frac{\mathrm{P}}{\mathrm{IV}}$ $\cos \phi=\frac{63}{3 \times 210}$ $\cos \phi=0.10$
GUJCET 2017
Alternating Current
155334
In an $\mathrm{AC}$ circuit the potential difference and current $i$ are represented respectively by $V=$ $100 \sin (100 \mathrm{t})$ volt, and $I=100 \sin \left(100 t+\frac{\pi}{3}\right)$ milli ampere. The power in the circuit is
155331
An inductor and a resistor are connected in series to an ac source of variable frequency. the power factor of the circuit is $\frac{\sqrt{3}}{2}$. If the frequency of the ac is increased by $200 \%$, the power factor of the circuit is
1 0.8
2 0.9
3 0.7
4 0.5
Explanation:
D According to question, Given, power factor $(\cos \phi)=\frac{\sqrt{3}}{2}$ $\mathrm{f}=50 \mathrm{~Hz}$ $\mathrm{f}^{\prime}=\mathrm{f}+200 \% \text { of } \mathrm{f}$ $\mathrm{f}^{\prime}=\mathrm{f}+\frac{200}{100} \times \mathrm{f}$ $\mathrm{f}^{\prime}=3 \mathrm{f}$ For R, L circuit $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+X_{L}^{2}}}$ $\cos ^{2} \phi=\frac{R^{2}}{R^{2}+X_{L}^{2}}$ $\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{R^{2}}{R^{2}+X_{L}^{2}}$ $\frac{3}{4}=\frac{R^{2}}{R^{2}+X_{L}^{2}}$ $3 R^{2}+3 X_{L}^{2}=4 R^{2}$ $R^{2}=3 \times X_{L}^{2}$ $X_{L}=\frac{R}{\sqrt{3}}$ $\because \quad \mathrm{X}_{\mathrm{L}} \propto \mathrm{f}$ $\therefore \quad$ After change in frequency, $\mathrm{X}_{\mathrm{L}}^{\prime}=3 \mathrm{~L}$ $\mathrm{X}_{\mathrm{L}}^{\prime}=3 \times \frac{\mathrm{R}}{\sqrt{3}}$ $\therefore \quad \cos \phi^{\prime} =\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{\prime 2}}}$ $\cos \phi^{\prime} =\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+(\sqrt{3} \mathrm{R})^{2}}}$ When the frequency of the applied ac is $50 \mathrm{~Hz}$, $\cos \phi^{\prime} =\frac{\mathrm{R}}{\sqrt{4} \mathrm{R}^{2}}$ $\cos \phi^{\prime} =\frac{\mathrm{R}}{2 \mathrm{R}}$ $\cos \phi^{\prime} =0.5$
AP EAMCET-24.04.2017
Alternating Current
155332
What is power dissipation in an A.C. circuit in which voltage and current are given by $V=300 \sin (\omega t+\pi / 2), I=5 \sin \omega t$ ?
155333
In an AC circuit, current is $3 \mathrm{~A}$ and voltage 210 $\mathrm{V}$ and power is $63 \mathrm{~W}$. The power factor is
1 0.11
2 0.09
3 0.08
4 0.10
Explanation:
D Given, $\mathrm{I}=3 \mathrm{~A}$ $\mathrm{~V}=210 \mathrm{~V}$ $\mathrm{P}=63 \mathrm{~W}$ $\because$ Power drawn by the circuit is - $\mathrm{P}=\mathrm{IV} \cos \phi$ Power factor, $(\cos \phi)=\frac{\mathrm{P}}{\mathrm{IV}}$ $\cos \phi=\frac{63}{3 \times 210}$ $\cos \phi=0.10$
GUJCET 2017
Alternating Current
155334
In an $\mathrm{AC}$ circuit the potential difference and current $i$ are represented respectively by $V=$ $100 \sin (100 \mathrm{t})$ volt, and $I=100 \sin \left(100 t+\frac{\pi}{3}\right)$ milli ampere. The power in the circuit is
155331
An inductor and a resistor are connected in series to an ac source of variable frequency. the power factor of the circuit is $\frac{\sqrt{3}}{2}$. If the frequency of the ac is increased by $200 \%$, the power factor of the circuit is
1 0.8
2 0.9
3 0.7
4 0.5
Explanation:
D According to question, Given, power factor $(\cos \phi)=\frac{\sqrt{3}}{2}$ $\mathrm{f}=50 \mathrm{~Hz}$ $\mathrm{f}^{\prime}=\mathrm{f}+200 \% \text { of } \mathrm{f}$ $\mathrm{f}^{\prime}=\mathrm{f}+\frac{200}{100} \times \mathrm{f}$ $\mathrm{f}^{\prime}=3 \mathrm{f}$ For R, L circuit $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+X_{L}^{2}}}$ $\cos ^{2} \phi=\frac{R^{2}}{R^{2}+X_{L}^{2}}$ $\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{R^{2}}{R^{2}+X_{L}^{2}}$ $\frac{3}{4}=\frac{R^{2}}{R^{2}+X_{L}^{2}}$ $3 R^{2}+3 X_{L}^{2}=4 R^{2}$ $R^{2}=3 \times X_{L}^{2}$ $X_{L}=\frac{R}{\sqrt{3}}$ $\because \quad \mathrm{X}_{\mathrm{L}} \propto \mathrm{f}$ $\therefore \quad$ After change in frequency, $\mathrm{X}_{\mathrm{L}}^{\prime}=3 \mathrm{~L}$ $\mathrm{X}_{\mathrm{L}}^{\prime}=3 \times \frac{\mathrm{R}}{\sqrt{3}}$ $\therefore \quad \cos \phi^{\prime} =\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{\prime 2}}}$ $\cos \phi^{\prime} =\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+(\sqrt{3} \mathrm{R})^{2}}}$ When the frequency of the applied ac is $50 \mathrm{~Hz}$, $\cos \phi^{\prime} =\frac{\mathrm{R}}{\sqrt{4} \mathrm{R}^{2}}$ $\cos \phi^{\prime} =\frac{\mathrm{R}}{2 \mathrm{R}}$ $\cos \phi^{\prime} =0.5$
AP EAMCET-24.04.2017
Alternating Current
155332
What is power dissipation in an A.C. circuit in which voltage and current are given by $V=300 \sin (\omega t+\pi / 2), I=5 \sin \omega t$ ?
155333
In an AC circuit, current is $3 \mathrm{~A}$ and voltage 210 $\mathrm{V}$ and power is $63 \mathrm{~W}$. The power factor is
1 0.11
2 0.09
3 0.08
4 0.10
Explanation:
D Given, $\mathrm{I}=3 \mathrm{~A}$ $\mathrm{~V}=210 \mathrm{~V}$ $\mathrm{P}=63 \mathrm{~W}$ $\because$ Power drawn by the circuit is - $\mathrm{P}=\mathrm{IV} \cos \phi$ Power factor, $(\cos \phi)=\frac{\mathrm{P}}{\mathrm{IV}}$ $\cos \phi=\frac{63}{3 \times 210}$ $\cos \phi=0.10$
GUJCET 2017
Alternating Current
155334
In an $\mathrm{AC}$ circuit the potential difference and current $i$ are represented respectively by $V=$ $100 \sin (100 \mathrm{t})$ volt, and $I=100 \sin \left(100 t+\frac{\pi}{3}\right)$ milli ampere. The power in the circuit is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155331
An inductor and a resistor are connected in series to an ac source of variable frequency. the power factor of the circuit is $\frac{\sqrt{3}}{2}$. If the frequency of the ac is increased by $200 \%$, the power factor of the circuit is
1 0.8
2 0.9
3 0.7
4 0.5
Explanation:
D According to question, Given, power factor $(\cos \phi)=\frac{\sqrt{3}}{2}$ $\mathrm{f}=50 \mathrm{~Hz}$ $\mathrm{f}^{\prime}=\mathrm{f}+200 \% \text { of } \mathrm{f}$ $\mathrm{f}^{\prime}=\mathrm{f}+\frac{200}{100} \times \mathrm{f}$ $\mathrm{f}^{\prime}=3 \mathrm{f}$ For R, L circuit $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+X_{L}^{2}}}$ $\cos ^{2} \phi=\frac{R^{2}}{R^{2}+X_{L}^{2}}$ $\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{R^{2}}{R^{2}+X_{L}^{2}}$ $\frac{3}{4}=\frac{R^{2}}{R^{2}+X_{L}^{2}}$ $3 R^{2}+3 X_{L}^{2}=4 R^{2}$ $R^{2}=3 \times X_{L}^{2}$ $X_{L}=\frac{R}{\sqrt{3}}$ $\because \quad \mathrm{X}_{\mathrm{L}} \propto \mathrm{f}$ $\therefore \quad$ After change in frequency, $\mathrm{X}_{\mathrm{L}}^{\prime}=3 \mathrm{~L}$ $\mathrm{X}_{\mathrm{L}}^{\prime}=3 \times \frac{\mathrm{R}}{\sqrt{3}}$ $\therefore \quad \cos \phi^{\prime} =\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{\prime 2}}}$ $\cos \phi^{\prime} =\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+(\sqrt{3} \mathrm{R})^{2}}}$ When the frequency of the applied ac is $50 \mathrm{~Hz}$, $\cos \phi^{\prime} =\frac{\mathrm{R}}{\sqrt{4} \mathrm{R}^{2}}$ $\cos \phi^{\prime} =\frac{\mathrm{R}}{2 \mathrm{R}}$ $\cos \phi^{\prime} =0.5$
AP EAMCET-24.04.2017
Alternating Current
155332
What is power dissipation in an A.C. circuit in which voltage and current are given by $V=300 \sin (\omega t+\pi / 2), I=5 \sin \omega t$ ?
155333
In an AC circuit, current is $3 \mathrm{~A}$ and voltage 210 $\mathrm{V}$ and power is $63 \mathrm{~W}$. The power factor is
1 0.11
2 0.09
3 0.08
4 0.10
Explanation:
D Given, $\mathrm{I}=3 \mathrm{~A}$ $\mathrm{~V}=210 \mathrm{~V}$ $\mathrm{P}=63 \mathrm{~W}$ $\because$ Power drawn by the circuit is - $\mathrm{P}=\mathrm{IV} \cos \phi$ Power factor, $(\cos \phi)=\frac{\mathrm{P}}{\mathrm{IV}}$ $\cos \phi=\frac{63}{3 \times 210}$ $\cos \phi=0.10$
GUJCET 2017
Alternating Current
155334
In an $\mathrm{AC}$ circuit the potential difference and current $i$ are represented respectively by $V=$ $100 \sin (100 \mathrm{t})$ volt, and $I=100 \sin \left(100 t+\frac{\pi}{3}\right)$ milli ampere. The power in the circuit is
