155311
In series $L C R$ circuit, resistance is $18 \Omega$ and impedance is $33 \Omega$. An r.m.s. voltage of $220 \mathrm{~V}$ is applied across the circuit. The true power consumed in a.c. circuit is
155312
Two electric circuit $A$ and $B$ are shown in the figure. The ratio of power factor of circuit $B$ to that of circuit $A$ is
1 $\sqrt{3}: 2$
2 $\sqrt{2}: 1$
3 $2: 3$
4 $4: 3$
Explanation:
B Given, $X_{L}=3 R, X_{C}=R$ Now, the power factor is $\mathrm{P}=\frac{\mathrm{I}^{2} \mathrm{R}}{\mathrm{I}^{2} \mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{Z}}$ Then, $\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}=\frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{\sqrt{(\mathrm{R})^{2}+\left(\mathrm{X}_{\mathrm{L}_{\mathrm{A}}}-\mathrm{X}_{\left.\mathrm{C}_{\mathrm{A}}\right)^{2}}\right.}}{\sqrt{(\mathrm{R})^{2}+\left(\mathrm{X}_{\mathrm{L}_{\mathrm{B}}}-\mathrm{X}_{\left.\mathrm{C}_{\mathrm{B}}\right)^{2}}\right.}}$ $\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}=\frac{\sqrt{\mathrm{R}^{2}+(3 \mathrm{R})^{2}}}{\sqrt{\mathrm{R}^{2}+(3 \mathrm{R}-\mathrm{R})^{2}}}=\frac{\sqrt{10 \mathrm{R}^{2}}}{\sqrt{5 \mathrm{R}^{2}}}=\frac{\sqrt{2}}{1}$ Hence, The ratio of power factor of circuit $B$ to that of circuit $A$ is $\sqrt{2}: 1$
AP EAMCET-07.10.2020
Alternating Current
155313
In LCR series circuit, an alternating e.m.f. 'e' and current ' $i$ ' are given by the equations $e=$ $100 \sin (100 t)$ volt, $\mathrm{i}=100 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathbf{m A} \text {. }$ The average power dissipated in the circuit will be
1 $100 \mathrm{~W}$
2 $10 \mathrm{~W}$
3 $5 \mathrm{~W}$
4 $2.5 \mathrm{~W}$
Explanation:
D Given that, $e=100 \sin (100 t) \text { Volt }$ $i=100 \sin (100 t+\pi / 3) m A$ Comparing with standard equation, $\mathrm{e}=\mathrm{e}_{0} \sin \omega \mathrm{t}, \mathrm{I}=\mathrm{I}_{0} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{I}_{0}=100 \mathrm{~mA}, \mathrm{e}_{0}=100 \mathrm{Volt}$ $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}} \mathrm{~mA}$ $\mathrm{e}_{\mathrm{rms}}=\frac{\mathrm{e}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}, \phi=\frac{\pi}{3}$ The average power, $P_{\text {avg }}=I_{r m s} \times e_{r m s} \times \cos \phi$ $=\frac{100}{\sqrt{2}} \times \frac{100}{\sqrt{2}} \times \cos 60^{\circ} \times 10^{-3}=2.5 \mathrm{~W}$
MHT-CET 2020
Alternating Current
155314
In an $\mathrm{AC}$ circuit, $\mathrm{V}$ and $\mathrm{I}$ are given by $V=150 \sin (150 t)$ volt and $I=150 \mathrm{sin}$ $\left(150 t+\frac{\pi}{3}\right)$ ampere. The power dissipated in the circuit is
1 $106 \mathrm{~W}$
2 $150 \mathrm{~W}$
3 $5625 \mathrm{~W}$
4 zero
Explanation:
C Given, $\mathrm{V}=150 \sin (150 \mathrm{t}) \text { Volt }$ Comparing with standard equation of AC voltage, $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ $\mathrm{V}_{0}=150$ And, $I=150 \sin \left(150 t+\frac{\pi}{3}\right)$ Comparing with standard equation of AC current, $I=I_{0} \sin (\omega t+\phi)$ $I_{0}=150 A$ Phase angle, $\phi=\frac{\pi}{3}=60^{\circ}$ We know that, The power dissipate in ac circuit is $\mathrm{P}=\frac{1}{2} \mathrm{~V}_{0} \mathrm{I}_{0} \cos \phi$ $\mathrm{P}=\frac{1}{2} \times 150 \times 150 \cos 60^{\circ}$ $\mathrm{P}=\frac{1}{2} \times 150 \times 150 \times \frac{1}{2}$ $\mathrm{P}=5625 \mathrm{~W}$ Hence, the power dissipated in the circuit is $5625 \mathrm{~W}$.
Karnataka CET-2011
Alternating Current
155316
A light bulb is rated at $110 \mathrm{~W}$ for a $220 \mathrm{~V}$ supply. The resistance of the bulb is
1 $440 \Omega$
2 $220 \Omega$
3 $55 \Omega$
4 $110 \Omega$
Explanation:
A Given, $\mathrm{P}=110 \mathrm{~W}$ $\mathrm{~V}=220 \mathrm{~V}$ We know that, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{110}=440 \Omega$
155311
In series $L C R$ circuit, resistance is $18 \Omega$ and impedance is $33 \Omega$. An r.m.s. voltage of $220 \mathrm{~V}$ is applied across the circuit. The true power consumed in a.c. circuit is
155312
Two electric circuit $A$ and $B$ are shown in the figure. The ratio of power factor of circuit $B$ to that of circuit $A$ is
1 $\sqrt{3}: 2$
2 $\sqrt{2}: 1$
3 $2: 3$
4 $4: 3$
Explanation:
B Given, $X_{L}=3 R, X_{C}=R$ Now, the power factor is $\mathrm{P}=\frac{\mathrm{I}^{2} \mathrm{R}}{\mathrm{I}^{2} \mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{Z}}$ Then, $\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}=\frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{\sqrt{(\mathrm{R})^{2}+\left(\mathrm{X}_{\mathrm{L}_{\mathrm{A}}}-\mathrm{X}_{\left.\mathrm{C}_{\mathrm{A}}\right)^{2}}\right.}}{\sqrt{(\mathrm{R})^{2}+\left(\mathrm{X}_{\mathrm{L}_{\mathrm{B}}}-\mathrm{X}_{\left.\mathrm{C}_{\mathrm{B}}\right)^{2}}\right.}}$ $\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}=\frac{\sqrt{\mathrm{R}^{2}+(3 \mathrm{R})^{2}}}{\sqrt{\mathrm{R}^{2}+(3 \mathrm{R}-\mathrm{R})^{2}}}=\frac{\sqrt{10 \mathrm{R}^{2}}}{\sqrt{5 \mathrm{R}^{2}}}=\frac{\sqrt{2}}{1}$ Hence, The ratio of power factor of circuit $B$ to that of circuit $A$ is $\sqrt{2}: 1$
AP EAMCET-07.10.2020
Alternating Current
155313
In LCR series circuit, an alternating e.m.f. 'e' and current ' $i$ ' are given by the equations $e=$ $100 \sin (100 t)$ volt, $\mathrm{i}=100 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathbf{m A} \text {. }$ The average power dissipated in the circuit will be
1 $100 \mathrm{~W}$
2 $10 \mathrm{~W}$
3 $5 \mathrm{~W}$
4 $2.5 \mathrm{~W}$
Explanation:
D Given that, $e=100 \sin (100 t) \text { Volt }$ $i=100 \sin (100 t+\pi / 3) m A$ Comparing with standard equation, $\mathrm{e}=\mathrm{e}_{0} \sin \omega \mathrm{t}, \mathrm{I}=\mathrm{I}_{0} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{I}_{0}=100 \mathrm{~mA}, \mathrm{e}_{0}=100 \mathrm{Volt}$ $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}} \mathrm{~mA}$ $\mathrm{e}_{\mathrm{rms}}=\frac{\mathrm{e}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}, \phi=\frac{\pi}{3}$ The average power, $P_{\text {avg }}=I_{r m s} \times e_{r m s} \times \cos \phi$ $=\frac{100}{\sqrt{2}} \times \frac{100}{\sqrt{2}} \times \cos 60^{\circ} \times 10^{-3}=2.5 \mathrm{~W}$
MHT-CET 2020
Alternating Current
155314
In an $\mathrm{AC}$ circuit, $\mathrm{V}$ and $\mathrm{I}$ are given by $V=150 \sin (150 t)$ volt and $I=150 \mathrm{sin}$ $\left(150 t+\frac{\pi}{3}\right)$ ampere. The power dissipated in the circuit is
1 $106 \mathrm{~W}$
2 $150 \mathrm{~W}$
3 $5625 \mathrm{~W}$
4 zero
Explanation:
C Given, $\mathrm{V}=150 \sin (150 \mathrm{t}) \text { Volt }$ Comparing with standard equation of AC voltage, $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ $\mathrm{V}_{0}=150$ And, $I=150 \sin \left(150 t+\frac{\pi}{3}\right)$ Comparing with standard equation of AC current, $I=I_{0} \sin (\omega t+\phi)$ $I_{0}=150 A$ Phase angle, $\phi=\frac{\pi}{3}=60^{\circ}$ We know that, The power dissipate in ac circuit is $\mathrm{P}=\frac{1}{2} \mathrm{~V}_{0} \mathrm{I}_{0} \cos \phi$ $\mathrm{P}=\frac{1}{2} \times 150 \times 150 \cos 60^{\circ}$ $\mathrm{P}=\frac{1}{2} \times 150 \times 150 \times \frac{1}{2}$ $\mathrm{P}=5625 \mathrm{~W}$ Hence, the power dissipated in the circuit is $5625 \mathrm{~W}$.
Karnataka CET-2011
Alternating Current
155316
A light bulb is rated at $110 \mathrm{~W}$ for a $220 \mathrm{~V}$ supply. The resistance of the bulb is
1 $440 \Omega$
2 $220 \Omega$
3 $55 \Omega$
4 $110 \Omega$
Explanation:
A Given, $\mathrm{P}=110 \mathrm{~W}$ $\mathrm{~V}=220 \mathrm{~V}$ We know that, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{110}=440 \Omega$
155311
In series $L C R$ circuit, resistance is $18 \Omega$ and impedance is $33 \Omega$. An r.m.s. voltage of $220 \mathrm{~V}$ is applied across the circuit. The true power consumed in a.c. circuit is
155312
Two electric circuit $A$ and $B$ are shown in the figure. The ratio of power factor of circuit $B$ to that of circuit $A$ is
1 $\sqrt{3}: 2$
2 $\sqrt{2}: 1$
3 $2: 3$
4 $4: 3$
Explanation:
B Given, $X_{L}=3 R, X_{C}=R$ Now, the power factor is $\mathrm{P}=\frac{\mathrm{I}^{2} \mathrm{R}}{\mathrm{I}^{2} \mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{Z}}$ Then, $\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}=\frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{\sqrt{(\mathrm{R})^{2}+\left(\mathrm{X}_{\mathrm{L}_{\mathrm{A}}}-\mathrm{X}_{\left.\mathrm{C}_{\mathrm{A}}\right)^{2}}\right.}}{\sqrt{(\mathrm{R})^{2}+\left(\mathrm{X}_{\mathrm{L}_{\mathrm{B}}}-\mathrm{X}_{\left.\mathrm{C}_{\mathrm{B}}\right)^{2}}\right.}}$ $\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}=\frac{\sqrt{\mathrm{R}^{2}+(3 \mathrm{R})^{2}}}{\sqrt{\mathrm{R}^{2}+(3 \mathrm{R}-\mathrm{R})^{2}}}=\frac{\sqrt{10 \mathrm{R}^{2}}}{\sqrt{5 \mathrm{R}^{2}}}=\frac{\sqrt{2}}{1}$ Hence, The ratio of power factor of circuit $B$ to that of circuit $A$ is $\sqrt{2}: 1$
AP EAMCET-07.10.2020
Alternating Current
155313
In LCR series circuit, an alternating e.m.f. 'e' and current ' $i$ ' are given by the equations $e=$ $100 \sin (100 t)$ volt, $\mathrm{i}=100 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathbf{m A} \text {. }$ The average power dissipated in the circuit will be
1 $100 \mathrm{~W}$
2 $10 \mathrm{~W}$
3 $5 \mathrm{~W}$
4 $2.5 \mathrm{~W}$
Explanation:
D Given that, $e=100 \sin (100 t) \text { Volt }$ $i=100 \sin (100 t+\pi / 3) m A$ Comparing with standard equation, $\mathrm{e}=\mathrm{e}_{0} \sin \omega \mathrm{t}, \mathrm{I}=\mathrm{I}_{0} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{I}_{0}=100 \mathrm{~mA}, \mathrm{e}_{0}=100 \mathrm{Volt}$ $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}} \mathrm{~mA}$ $\mathrm{e}_{\mathrm{rms}}=\frac{\mathrm{e}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}, \phi=\frac{\pi}{3}$ The average power, $P_{\text {avg }}=I_{r m s} \times e_{r m s} \times \cos \phi$ $=\frac{100}{\sqrt{2}} \times \frac{100}{\sqrt{2}} \times \cos 60^{\circ} \times 10^{-3}=2.5 \mathrm{~W}$
MHT-CET 2020
Alternating Current
155314
In an $\mathrm{AC}$ circuit, $\mathrm{V}$ and $\mathrm{I}$ are given by $V=150 \sin (150 t)$ volt and $I=150 \mathrm{sin}$ $\left(150 t+\frac{\pi}{3}\right)$ ampere. The power dissipated in the circuit is
1 $106 \mathrm{~W}$
2 $150 \mathrm{~W}$
3 $5625 \mathrm{~W}$
4 zero
Explanation:
C Given, $\mathrm{V}=150 \sin (150 \mathrm{t}) \text { Volt }$ Comparing with standard equation of AC voltage, $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ $\mathrm{V}_{0}=150$ And, $I=150 \sin \left(150 t+\frac{\pi}{3}\right)$ Comparing with standard equation of AC current, $I=I_{0} \sin (\omega t+\phi)$ $I_{0}=150 A$ Phase angle, $\phi=\frac{\pi}{3}=60^{\circ}$ We know that, The power dissipate in ac circuit is $\mathrm{P}=\frac{1}{2} \mathrm{~V}_{0} \mathrm{I}_{0} \cos \phi$ $\mathrm{P}=\frac{1}{2} \times 150 \times 150 \cos 60^{\circ}$ $\mathrm{P}=\frac{1}{2} \times 150 \times 150 \times \frac{1}{2}$ $\mathrm{P}=5625 \mathrm{~W}$ Hence, the power dissipated in the circuit is $5625 \mathrm{~W}$.
Karnataka CET-2011
Alternating Current
155316
A light bulb is rated at $110 \mathrm{~W}$ for a $220 \mathrm{~V}$ supply. The resistance of the bulb is
1 $440 \Omega$
2 $220 \Omega$
3 $55 \Omega$
4 $110 \Omega$
Explanation:
A Given, $\mathrm{P}=110 \mathrm{~W}$ $\mathrm{~V}=220 \mathrm{~V}$ We know that, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{110}=440 \Omega$
155311
In series $L C R$ circuit, resistance is $18 \Omega$ and impedance is $33 \Omega$. An r.m.s. voltage of $220 \mathrm{~V}$ is applied across the circuit. The true power consumed in a.c. circuit is
155312
Two electric circuit $A$ and $B$ are shown in the figure. The ratio of power factor of circuit $B$ to that of circuit $A$ is
1 $\sqrt{3}: 2$
2 $\sqrt{2}: 1$
3 $2: 3$
4 $4: 3$
Explanation:
B Given, $X_{L}=3 R, X_{C}=R$ Now, the power factor is $\mathrm{P}=\frac{\mathrm{I}^{2} \mathrm{R}}{\mathrm{I}^{2} \mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{Z}}$ Then, $\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}=\frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{\sqrt{(\mathrm{R})^{2}+\left(\mathrm{X}_{\mathrm{L}_{\mathrm{A}}}-\mathrm{X}_{\left.\mathrm{C}_{\mathrm{A}}\right)^{2}}\right.}}{\sqrt{(\mathrm{R})^{2}+\left(\mathrm{X}_{\mathrm{L}_{\mathrm{B}}}-\mathrm{X}_{\left.\mathrm{C}_{\mathrm{B}}\right)^{2}}\right.}}$ $\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}=\frac{\sqrt{\mathrm{R}^{2}+(3 \mathrm{R})^{2}}}{\sqrt{\mathrm{R}^{2}+(3 \mathrm{R}-\mathrm{R})^{2}}}=\frac{\sqrt{10 \mathrm{R}^{2}}}{\sqrt{5 \mathrm{R}^{2}}}=\frac{\sqrt{2}}{1}$ Hence, The ratio of power factor of circuit $B$ to that of circuit $A$ is $\sqrt{2}: 1$
AP EAMCET-07.10.2020
Alternating Current
155313
In LCR series circuit, an alternating e.m.f. 'e' and current ' $i$ ' are given by the equations $e=$ $100 \sin (100 t)$ volt, $\mathrm{i}=100 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathbf{m A} \text {. }$ The average power dissipated in the circuit will be
1 $100 \mathrm{~W}$
2 $10 \mathrm{~W}$
3 $5 \mathrm{~W}$
4 $2.5 \mathrm{~W}$
Explanation:
D Given that, $e=100 \sin (100 t) \text { Volt }$ $i=100 \sin (100 t+\pi / 3) m A$ Comparing with standard equation, $\mathrm{e}=\mathrm{e}_{0} \sin \omega \mathrm{t}, \mathrm{I}=\mathrm{I}_{0} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{I}_{0}=100 \mathrm{~mA}, \mathrm{e}_{0}=100 \mathrm{Volt}$ $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}} \mathrm{~mA}$ $\mathrm{e}_{\mathrm{rms}}=\frac{\mathrm{e}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}, \phi=\frac{\pi}{3}$ The average power, $P_{\text {avg }}=I_{r m s} \times e_{r m s} \times \cos \phi$ $=\frac{100}{\sqrt{2}} \times \frac{100}{\sqrt{2}} \times \cos 60^{\circ} \times 10^{-3}=2.5 \mathrm{~W}$
MHT-CET 2020
Alternating Current
155314
In an $\mathrm{AC}$ circuit, $\mathrm{V}$ and $\mathrm{I}$ are given by $V=150 \sin (150 t)$ volt and $I=150 \mathrm{sin}$ $\left(150 t+\frac{\pi}{3}\right)$ ampere. The power dissipated in the circuit is
1 $106 \mathrm{~W}$
2 $150 \mathrm{~W}$
3 $5625 \mathrm{~W}$
4 zero
Explanation:
C Given, $\mathrm{V}=150 \sin (150 \mathrm{t}) \text { Volt }$ Comparing with standard equation of AC voltage, $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ $\mathrm{V}_{0}=150$ And, $I=150 \sin \left(150 t+\frac{\pi}{3}\right)$ Comparing with standard equation of AC current, $I=I_{0} \sin (\omega t+\phi)$ $I_{0}=150 A$ Phase angle, $\phi=\frac{\pi}{3}=60^{\circ}$ We know that, The power dissipate in ac circuit is $\mathrm{P}=\frac{1}{2} \mathrm{~V}_{0} \mathrm{I}_{0} \cos \phi$ $\mathrm{P}=\frac{1}{2} \times 150 \times 150 \cos 60^{\circ}$ $\mathrm{P}=\frac{1}{2} \times 150 \times 150 \times \frac{1}{2}$ $\mathrm{P}=5625 \mathrm{~W}$ Hence, the power dissipated in the circuit is $5625 \mathrm{~W}$.
Karnataka CET-2011
Alternating Current
155316
A light bulb is rated at $110 \mathrm{~W}$ for a $220 \mathrm{~V}$ supply. The resistance of the bulb is
1 $440 \Omega$
2 $220 \Omega$
3 $55 \Omega$
4 $110 \Omega$
Explanation:
A Given, $\mathrm{P}=110 \mathrm{~W}$ $\mathrm{~V}=220 \mathrm{~V}$ We know that, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{110}=440 \Omega$
155311
In series $L C R$ circuit, resistance is $18 \Omega$ and impedance is $33 \Omega$. An r.m.s. voltage of $220 \mathrm{~V}$ is applied across the circuit. The true power consumed in a.c. circuit is
155312
Two electric circuit $A$ and $B$ are shown in the figure. The ratio of power factor of circuit $B$ to that of circuit $A$ is
1 $\sqrt{3}: 2$
2 $\sqrt{2}: 1$
3 $2: 3$
4 $4: 3$
Explanation:
B Given, $X_{L}=3 R, X_{C}=R$ Now, the power factor is $\mathrm{P}=\frac{\mathrm{I}^{2} \mathrm{R}}{\mathrm{I}^{2} \mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{Z}}$ Then, $\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}=\frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{\sqrt{(\mathrm{R})^{2}+\left(\mathrm{X}_{\mathrm{L}_{\mathrm{A}}}-\mathrm{X}_{\left.\mathrm{C}_{\mathrm{A}}\right)^{2}}\right.}}{\sqrt{(\mathrm{R})^{2}+\left(\mathrm{X}_{\mathrm{L}_{\mathrm{B}}}-\mathrm{X}_{\left.\mathrm{C}_{\mathrm{B}}\right)^{2}}\right.}}$ $\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{A}}}=\frac{\sqrt{\mathrm{R}^{2}+(3 \mathrm{R})^{2}}}{\sqrt{\mathrm{R}^{2}+(3 \mathrm{R}-\mathrm{R})^{2}}}=\frac{\sqrt{10 \mathrm{R}^{2}}}{\sqrt{5 \mathrm{R}^{2}}}=\frac{\sqrt{2}}{1}$ Hence, The ratio of power factor of circuit $B$ to that of circuit $A$ is $\sqrt{2}: 1$
AP EAMCET-07.10.2020
Alternating Current
155313
In LCR series circuit, an alternating e.m.f. 'e' and current ' $i$ ' are given by the equations $e=$ $100 \sin (100 t)$ volt, $\mathrm{i}=100 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathbf{m A} \text {. }$ The average power dissipated in the circuit will be
1 $100 \mathrm{~W}$
2 $10 \mathrm{~W}$
3 $5 \mathrm{~W}$
4 $2.5 \mathrm{~W}$
Explanation:
D Given that, $e=100 \sin (100 t) \text { Volt }$ $i=100 \sin (100 t+\pi / 3) m A$ Comparing with standard equation, $\mathrm{e}=\mathrm{e}_{0} \sin \omega \mathrm{t}, \mathrm{I}=\mathrm{I}_{0} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{I}_{0}=100 \mathrm{~mA}, \mathrm{e}_{0}=100 \mathrm{Volt}$ $\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{I}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}} \mathrm{~mA}$ $\mathrm{e}_{\mathrm{rms}}=\frac{\mathrm{e}_{0}}{\sqrt{2}}=\frac{100}{\sqrt{2}}, \phi=\frac{\pi}{3}$ The average power, $P_{\text {avg }}=I_{r m s} \times e_{r m s} \times \cos \phi$ $=\frac{100}{\sqrt{2}} \times \frac{100}{\sqrt{2}} \times \cos 60^{\circ} \times 10^{-3}=2.5 \mathrm{~W}$
MHT-CET 2020
Alternating Current
155314
In an $\mathrm{AC}$ circuit, $\mathrm{V}$ and $\mathrm{I}$ are given by $V=150 \sin (150 t)$ volt and $I=150 \mathrm{sin}$ $\left(150 t+\frac{\pi}{3}\right)$ ampere. The power dissipated in the circuit is
1 $106 \mathrm{~W}$
2 $150 \mathrm{~W}$
3 $5625 \mathrm{~W}$
4 zero
Explanation:
C Given, $\mathrm{V}=150 \sin (150 \mathrm{t}) \text { Volt }$ Comparing with standard equation of AC voltage, $\mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t}$ $\mathrm{V}_{0}=150$ And, $I=150 \sin \left(150 t+\frac{\pi}{3}\right)$ Comparing with standard equation of AC current, $I=I_{0} \sin (\omega t+\phi)$ $I_{0}=150 A$ Phase angle, $\phi=\frac{\pi}{3}=60^{\circ}$ We know that, The power dissipate in ac circuit is $\mathrm{P}=\frac{1}{2} \mathrm{~V}_{0} \mathrm{I}_{0} \cos \phi$ $\mathrm{P}=\frac{1}{2} \times 150 \times 150 \cos 60^{\circ}$ $\mathrm{P}=\frac{1}{2} \times 150 \times 150 \times \frac{1}{2}$ $\mathrm{P}=5625 \mathrm{~W}$ Hence, the power dissipated in the circuit is $5625 \mathrm{~W}$.
Karnataka CET-2011
Alternating Current
155316
A light bulb is rated at $110 \mathrm{~W}$ for a $220 \mathrm{~V}$ supply. The resistance of the bulb is
1 $440 \Omega$
2 $220 \Omega$
3 $55 \Omega$
4 $110 \Omega$
Explanation:
A Given, $\mathrm{P}=110 \mathrm{~W}$ $\mathrm{~V}=220 \mathrm{~V}$ We know that, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ $\mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(220)^{2}}{110}=440 \Omega$
