155335
A lamp consumes only $50 \%$ of maximum power applied in an A.C. circuit. What will be the phase difference between applied voltage and circuit current?
1 $\pi / 6 \mathrm{rad}$
2 $\pi / 3 \mathrm{rad}$
3 $\pi / 4 \mathrm{rad}$
4 $\pi / 2 \mathrm{rad}$
Explanation:
B Given, $\mathrm{P}=\frac{\mathrm{P}_{\mathrm{P}}}{2}$ We know that, $\mathrm{P}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\mathrm{rms}} \cos \phi$ Where, $V_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{p}}}{\sqrt{2}}$ and $\mathrm{R}_{\mathrm{rms}}=\frac{\mathrm{I}_{\mathrm{p}}}{\sqrt{2}}$ So, $\quad \mathrm{P}=\frac{\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}}{2} \cos \phi$ $\mathrm{P}=\mathrm{P}_{\mathrm{P}} \cos \phi$ $\frac{\mathrm{P}_{\mathrm{P}}}{2}=\mathrm{P}_{\mathrm{P}} \cos \phi$ Therefore, we get- $\phi=\cos ^{-1} \frac{1}{2}$ $\phi=\frac{\pi}{3}$ Thus, the phase difference between the applied voltage and circuit current when a lamp consumer only $50 \%$ of peak power in an AC circuit is $\pi / 3$.
GUJCET 2014
Alternating Current
155336
Amount of heat produced per second in calories when a bulb of $50 \mathrm{~W}, 200 \mathrm{~V}$ glows (assuming that only $20 \%$ of the electric energy is converted into light) ( $\mathrm{J}=4.2 \mathrm{~J} / \mathrm{cal}$ )
1 $40 \mathrm{cal} / \mathrm{s}$
2 $28 \mathrm{cal} / \mathrm{s}$
3 $18.22 \mathrm{cal} / \mathrm{s}$
4 $9.52 \mathrm{cal} / \mathrm{s}$
Explanation:
D Given, power of bulb $=50 \mathrm{~W}$, Time taken $=1 \mathrm{~s}$ Electric energy consumed per second $=50 \mathrm{~W}$ Heat produced $=100 \%-20 \%=80 \%$ $=\frac{80}{100} \times 50=40 \mathrm{~J}$ We know that, $1 \mathrm{cal}=4.2 \mathrm{~J}$ $1 \mathrm{~J}= \frac{1}{4.2} \mathrm{cal}$ $=\frac{40}{4.2} \mathrm{cal} / \mathrm{sec}=9.52 \mathrm{cal} / \mathrm{sec}$
CG PET -2016
Alternating Current
155338
A coil has resistance $30 \mathrm{ohm}$ and inductive reactance $20 \mathrm{ohm}$ at $50 \mathrm{~Hz}$ frequency. If an ac source, of 200 volt, $100 \mathrm{~Hz}$, is connected across the coil, the current in the coil will be
155339
The impedance of a circuit consists of $3 \Omega$ resistance and $4 \Omega$ reactance. The power factor of the circuit is
1 0.4
2 0.6
3 0.8
4 1.0
Explanation:
B Given, $\mathrm{R}=3 \Omega$ $\mathrm{X}_{\mathrm{L}}=4 \Omega$ We know that, impedance of the circuit - $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}=\sqrt{3^{2}+4^{2}}$ $\mathrm{Z}=5 \Omega$ The power factor is, $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{3}{5}$ $\cos \phi=0.6$
155335
A lamp consumes only $50 \%$ of maximum power applied in an A.C. circuit. What will be the phase difference between applied voltage and circuit current?
1 $\pi / 6 \mathrm{rad}$
2 $\pi / 3 \mathrm{rad}$
3 $\pi / 4 \mathrm{rad}$
4 $\pi / 2 \mathrm{rad}$
Explanation:
B Given, $\mathrm{P}=\frac{\mathrm{P}_{\mathrm{P}}}{2}$ We know that, $\mathrm{P}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\mathrm{rms}} \cos \phi$ Where, $V_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{p}}}{\sqrt{2}}$ and $\mathrm{R}_{\mathrm{rms}}=\frac{\mathrm{I}_{\mathrm{p}}}{\sqrt{2}}$ So, $\quad \mathrm{P}=\frac{\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}}{2} \cos \phi$ $\mathrm{P}=\mathrm{P}_{\mathrm{P}} \cos \phi$ $\frac{\mathrm{P}_{\mathrm{P}}}{2}=\mathrm{P}_{\mathrm{P}} \cos \phi$ Therefore, we get- $\phi=\cos ^{-1} \frac{1}{2}$ $\phi=\frac{\pi}{3}$ Thus, the phase difference between the applied voltage and circuit current when a lamp consumer only $50 \%$ of peak power in an AC circuit is $\pi / 3$.
GUJCET 2014
Alternating Current
155336
Amount of heat produced per second in calories when a bulb of $50 \mathrm{~W}, 200 \mathrm{~V}$ glows (assuming that only $20 \%$ of the electric energy is converted into light) ( $\mathrm{J}=4.2 \mathrm{~J} / \mathrm{cal}$ )
1 $40 \mathrm{cal} / \mathrm{s}$
2 $28 \mathrm{cal} / \mathrm{s}$
3 $18.22 \mathrm{cal} / \mathrm{s}$
4 $9.52 \mathrm{cal} / \mathrm{s}$
Explanation:
D Given, power of bulb $=50 \mathrm{~W}$, Time taken $=1 \mathrm{~s}$ Electric energy consumed per second $=50 \mathrm{~W}$ Heat produced $=100 \%-20 \%=80 \%$ $=\frac{80}{100} \times 50=40 \mathrm{~J}$ We know that, $1 \mathrm{cal}=4.2 \mathrm{~J}$ $1 \mathrm{~J}= \frac{1}{4.2} \mathrm{cal}$ $=\frac{40}{4.2} \mathrm{cal} / \mathrm{sec}=9.52 \mathrm{cal} / \mathrm{sec}$
CG PET -2016
Alternating Current
155338
A coil has resistance $30 \mathrm{ohm}$ and inductive reactance $20 \mathrm{ohm}$ at $50 \mathrm{~Hz}$ frequency. If an ac source, of 200 volt, $100 \mathrm{~Hz}$, is connected across the coil, the current in the coil will be
155339
The impedance of a circuit consists of $3 \Omega$ resistance and $4 \Omega$ reactance. The power factor of the circuit is
1 0.4
2 0.6
3 0.8
4 1.0
Explanation:
B Given, $\mathrm{R}=3 \Omega$ $\mathrm{X}_{\mathrm{L}}=4 \Omega$ We know that, impedance of the circuit - $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}=\sqrt{3^{2}+4^{2}}$ $\mathrm{Z}=5 \Omega$ The power factor is, $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{3}{5}$ $\cos \phi=0.6$
155335
A lamp consumes only $50 \%$ of maximum power applied in an A.C. circuit. What will be the phase difference between applied voltage and circuit current?
1 $\pi / 6 \mathrm{rad}$
2 $\pi / 3 \mathrm{rad}$
3 $\pi / 4 \mathrm{rad}$
4 $\pi / 2 \mathrm{rad}$
Explanation:
B Given, $\mathrm{P}=\frac{\mathrm{P}_{\mathrm{P}}}{2}$ We know that, $\mathrm{P}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\mathrm{rms}} \cos \phi$ Where, $V_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{p}}}{\sqrt{2}}$ and $\mathrm{R}_{\mathrm{rms}}=\frac{\mathrm{I}_{\mathrm{p}}}{\sqrt{2}}$ So, $\quad \mathrm{P}=\frac{\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}}{2} \cos \phi$ $\mathrm{P}=\mathrm{P}_{\mathrm{P}} \cos \phi$ $\frac{\mathrm{P}_{\mathrm{P}}}{2}=\mathrm{P}_{\mathrm{P}} \cos \phi$ Therefore, we get- $\phi=\cos ^{-1} \frac{1}{2}$ $\phi=\frac{\pi}{3}$ Thus, the phase difference between the applied voltage and circuit current when a lamp consumer only $50 \%$ of peak power in an AC circuit is $\pi / 3$.
GUJCET 2014
Alternating Current
155336
Amount of heat produced per second in calories when a bulb of $50 \mathrm{~W}, 200 \mathrm{~V}$ glows (assuming that only $20 \%$ of the electric energy is converted into light) ( $\mathrm{J}=4.2 \mathrm{~J} / \mathrm{cal}$ )
1 $40 \mathrm{cal} / \mathrm{s}$
2 $28 \mathrm{cal} / \mathrm{s}$
3 $18.22 \mathrm{cal} / \mathrm{s}$
4 $9.52 \mathrm{cal} / \mathrm{s}$
Explanation:
D Given, power of bulb $=50 \mathrm{~W}$, Time taken $=1 \mathrm{~s}$ Electric energy consumed per second $=50 \mathrm{~W}$ Heat produced $=100 \%-20 \%=80 \%$ $=\frac{80}{100} \times 50=40 \mathrm{~J}$ We know that, $1 \mathrm{cal}=4.2 \mathrm{~J}$ $1 \mathrm{~J}= \frac{1}{4.2} \mathrm{cal}$ $=\frac{40}{4.2} \mathrm{cal} / \mathrm{sec}=9.52 \mathrm{cal} / \mathrm{sec}$
CG PET -2016
Alternating Current
155338
A coil has resistance $30 \mathrm{ohm}$ and inductive reactance $20 \mathrm{ohm}$ at $50 \mathrm{~Hz}$ frequency. If an ac source, of 200 volt, $100 \mathrm{~Hz}$, is connected across the coil, the current in the coil will be
155339
The impedance of a circuit consists of $3 \Omega$ resistance and $4 \Omega$ reactance. The power factor of the circuit is
1 0.4
2 0.6
3 0.8
4 1.0
Explanation:
B Given, $\mathrm{R}=3 \Omega$ $\mathrm{X}_{\mathrm{L}}=4 \Omega$ We know that, impedance of the circuit - $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}=\sqrt{3^{2}+4^{2}}$ $\mathrm{Z}=5 \Omega$ The power factor is, $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{3}{5}$ $\cos \phi=0.6$
155335
A lamp consumes only $50 \%$ of maximum power applied in an A.C. circuit. What will be the phase difference between applied voltage and circuit current?
1 $\pi / 6 \mathrm{rad}$
2 $\pi / 3 \mathrm{rad}$
3 $\pi / 4 \mathrm{rad}$
4 $\pi / 2 \mathrm{rad}$
Explanation:
B Given, $\mathrm{P}=\frac{\mathrm{P}_{\mathrm{P}}}{2}$ We know that, $\mathrm{P}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\mathrm{rms}} \cos \phi$ Where, $V_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{p}}}{\sqrt{2}}$ and $\mathrm{R}_{\mathrm{rms}}=\frac{\mathrm{I}_{\mathrm{p}}}{\sqrt{2}}$ So, $\quad \mathrm{P}=\frac{\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}}{2} \cos \phi$ $\mathrm{P}=\mathrm{P}_{\mathrm{P}} \cos \phi$ $\frac{\mathrm{P}_{\mathrm{P}}}{2}=\mathrm{P}_{\mathrm{P}} \cos \phi$ Therefore, we get- $\phi=\cos ^{-1} \frac{1}{2}$ $\phi=\frac{\pi}{3}$ Thus, the phase difference between the applied voltage and circuit current when a lamp consumer only $50 \%$ of peak power in an AC circuit is $\pi / 3$.
GUJCET 2014
Alternating Current
155336
Amount of heat produced per second in calories when a bulb of $50 \mathrm{~W}, 200 \mathrm{~V}$ glows (assuming that only $20 \%$ of the electric energy is converted into light) ( $\mathrm{J}=4.2 \mathrm{~J} / \mathrm{cal}$ )
1 $40 \mathrm{cal} / \mathrm{s}$
2 $28 \mathrm{cal} / \mathrm{s}$
3 $18.22 \mathrm{cal} / \mathrm{s}$
4 $9.52 \mathrm{cal} / \mathrm{s}$
Explanation:
D Given, power of bulb $=50 \mathrm{~W}$, Time taken $=1 \mathrm{~s}$ Electric energy consumed per second $=50 \mathrm{~W}$ Heat produced $=100 \%-20 \%=80 \%$ $=\frac{80}{100} \times 50=40 \mathrm{~J}$ We know that, $1 \mathrm{cal}=4.2 \mathrm{~J}$ $1 \mathrm{~J}= \frac{1}{4.2} \mathrm{cal}$ $=\frac{40}{4.2} \mathrm{cal} / \mathrm{sec}=9.52 \mathrm{cal} / \mathrm{sec}$
CG PET -2016
Alternating Current
155338
A coil has resistance $30 \mathrm{ohm}$ and inductive reactance $20 \mathrm{ohm}$ at $50 \mathrm{~Hz}$ frequency. If an ac source, of 200 volt, $100 \mathrm{~Hz}$, is connected across the coil, the current in the coil will be
155339
The impedance of a circuit consists of $3 \Omega$ resistance and $4 \Omega$ reactance. The power factor of the circuit is
1 0.4
2 0.6
3 0.8
4 1.0
Explanation:
B Given, $\mathrm{R}=3 \Omega$ $\mathrm{X}_{\mathrm{L}}=4 \Omega$ We know that, impedance of the circuit - $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}=\sqrt{3^{2}+4^{2}}$ $\mathrm{Z}=5 \Omega$ The power factor is, $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{3}{5}$ $\cos \phi=0.6$
