155361
A coil of inductive reactance $31 \Omega$ has a resistance of $8 \Omega$. it is placed in series with a condenser of capacitive reactance $25 \Omega$. The combination is connected to an $\mathrm{AC}$ source of $110 \mathrm{~V}$. The power factor of the circuit is
1 0.56
2 0.64
3 0.80
4 0.33
Explanation:
C Given that, $\mathrm{X}_{\mathrm{L}}=31 \Omega, \mathrm{R}=8 \Omega \&, \mathrm{X}_{\mathrm{C}}=25 \Omega$ According to question, We know that, impedance of circuit - $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{8^{2}+(31-25)^{2}}$ $Z=\sqrt{64+36}$ $Z=10$ $\therefore$ Power factor, $\cos \phi=\frac{R}{Z}=\frac{8}{10}$ $\operatorname{Cos} \phi=0.8$
AIPMT- 2006
Alternating Current
155362
Power dissipated in an L-C-R series circuit connected to an $\mathrm{AC}$ source of emf $\varepsilon$ is
A According to question, Impedance of L-R-C series circuit $|Z|=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $|Z|=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$ $\cos \phi=\frac{R}{|Z|}$ Power dissipated, $\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\mathrm{P}=\varepsilon \times \frac{\varepsilon}{|\mathrm{Z}|} \times \frac{\mathrm{R}}{|\mathrm{Z}|}$ $\mathrm{P}=\frac{\varepsilon^{2} \mathrm{R}}{|\mathrm{Z}|^{2}}=\frac{\varepsilon^{2} \mathrm{R}}{\left(\sqrt{\mathrm{R}^{2}+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^{2}}\right)^{2}}$ $\mathrm{P}=\frac{\varepsilon^{2} \mathrm{R}}{\left[\mathrm{R}^{2}+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^{2}\right]}$
AIPMT- 2009
Alternating Current
155363
A resistance ' $R$ ' draws power ' $P$ ' when connected to an $\mathrm{AC}$ source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes ' $Z$ ' the power drawn will be
A Given that, Resistance $=\mathrm{R}$ Power $=\mathrm{P}$ We know that, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ When inductance is placed in series with resistance, Then, $\quad \mathrm{P}^{\prime}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\mathrm{rms}} \cos \phi$ In case of series $\mathrm{R}-\mathrm{L}$ circuit $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}$ $\mathrm{P}^{\prime}=\mathrm{V} \times\left(\frac{\mathrm{V}}{\mathrm{Z}}\right) \cdot \frac{\mathrm{R}}{\mathrm{Z}}$ $\mathrm{P}^{\prime}=\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}}$ Dividing equation (ii) by equation (i), we get- $\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}}}{\frac{\mathrm{V}^{2}}{\mathrm{R}}}=\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}} \times \frac{\mathrm{R}}{\mathrm{V}^{2}}$ $\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\mathrm{R}^{2}}{\mathrm{Z}^{2}}$ $\mathrm{P}^{\prime}=\mathrm{P}\left(\frac{\mathrm{R}}{\mathrm{Z}}\right)^{2}$
AIPMT- 2015
Alternating Current
155364
An inductor $20 \mathrm{mH}$, a capacitor $50 \mu \mathrm{F}$ and a resistor $40 \Omega$ are connected in series across a source of emf $V=10 \sin 340 t$. The power loss in $\mathrm{AC}$ circuit is
155361
A coil of inductive reactance $31 \Omega$ has a resistance of $8 \Omega$. it is placed in series with a condenser of capacitive reactance $25 \Omega$. The combination is connected to an $\mathrm{AC}$ source of $110 \mathrm{~V}$. The power factor of the circuit is
1 0.56
2 0.64
3 0.80
4 0.33
Explanation:
C Given that, $\mathrm{X}_{\mathrm{L}}=31 \Omega, \mathrm{R}=8 \Omega \&, \mathrm{X}_{\mathrm{C}}=25 \Omega$ According to question, We know that, impedance of circuit - $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{8^{2}+(31-25)^{2}}$ $Z=\sqrt{64+36}$ $Z=10$ $\therefore$ Power factor, $\cos \phi=\frac{R}{Z}=\frac{8}{10}$ $\operatorname{Cos} \phi=0.8$
AIPMT- 2006
Alternating Current
155362
Power dissipated in an L-C-R series circuit connected to an $\mathrm{AC}$ source of emf $\varepsilon$ is
A According to question, Impedance of L-R-C series circuit $|Z|=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $|Z|=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$ $\cos \phi=\frac{R}{|Z|}$ Power dissipated, $\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\mathrm{P}=\varepsilon \times \frac{\varepsilon}{|\mathrm{Z}|} \times \frac{\mathrm{R}}{|\mathrm{Z}|}$ $\mathrm{P}=\frac{\varepsilon^{2} \mathrm{R}}{|\mathrm{Z}|^{2}}=\frac{\varepsilon^{2} \mathrm{R}}{\left(\sqrt{\mathrm{R}^{2}+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^{2}}\right)^{2}}$ $\mathrm{P}=\frac{\varepsilon^{2} \mathrm{R}}{\left[\mathrm{R}^{2}+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^{2}\right]}$
AIPMT- 2009
Alternating Current
155363
A resistance ' $R$ ' draws power ' $P$ ' when connected to an $\mathrm{AC}$ source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes ' $Z$ ' the power drawn will be
A Given that, Resistance $=\mathrm{R}$ Power $=\mathrm{P}$ We know that, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ When inductance is placed in series with resistance, Then, $\quad \mathrm{P}^{\prime}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\mathrm{rms}} \cos \phi$ In case of series $\mathrm{R}-\mathrm{L}$ circuit $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}$ $\mathrm{P}^{\prime}=\mathrm{V} \times\left(\frac{\mathrm{V}}{\mathrm{Z}}\right) \cdot \frac{\mathrm{R}}{\mathrm{Z}}$ $\mathrm{P}^{\prime}=\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}}$ Dividing equation (ii) by equation (i), we get- $\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}}}{\frac{\mathrm{V}^{2}}{\mathrm{R}}}=\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}} \times \frac{\mathrm{R}}{\mathrm{V}^{2}}$ $\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\mathrm{R}^{2}}{\mathrm{Z}^{2}}$ $\mathrm{P}^{\prime}=\mathrm{P}\left(\frac{\mathrm{R}}{\mathrm{Z}}\right)^{2}$
AIPMT- 2015
Alternating Current
155364
An inductor $20 \mathrm{mH}$, a capacitor $50 \mu \mathrm{F}$ and a resistor $40 \Omega$ are connected in series across a source of emf $V=10 \sin 340 t$. The power loss in $\mathrm{AC}$ circuit is
155361
A coil of inductive reactance $31 \Omega$ has a resistance of $8 \Omega$. it is placed in series with a condenser of capacitive reactance $25 \Omega$. The combination is connected to an $\mathrm{AC}$ source of $110 \mathrm{~V}$. The power factor of the circuit is
1 0.56
2 0.64
3 0.80
4 0.33
Explanation:
C Given that, $\mathrm{X}_{\mathrm{L}}=31 \Omega, \mathrm{R}=8 \Omega \&, \mathrm{X}_{\mathrm{C}}=25 \Omega$ According to question, We know that, impedance of circuit - $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{8^{2}+(31-25)^{2}}$ $Z=\sqrt{64+36}$ $Z=10$ $\therefore$ Power factor, $\cos \phi=\frac{R}{Z}=\frac{8}{10}$ $\operatorname{Cos} \phi=0.8$
AIPMT- 2006
Alternating Current
155362
Power dissipated in an L-C-R series circuit connected to an $\mathrm{AC}$ source of emf $\varepsilon$ is
A According to question, Impedance of L-R-C series circuit $|Z|=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $|Z|=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$ $\cos \phi=\frac{R}{|Z|}$ Power dissipated, $\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\mathrm{P}=\varepsilon \times \frac{\varepsilon}{|\mathrm{Z}|} \times \frac{\mathrm{R}}{|\mathrm{Z}|}$ $\mathrm{P}=\frac{\varepsilon^{2} \mathrm{R}}{|\mathrm{Z}|^{2}}=\frac{\varepsilon^{2} \mathrm{R}}{\left(\sqrt{\mathrm{R}^{2}+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^{2}}\right)^{2}}$ $\mathrm{P}=\frac{\varepsilon^{2} \mathrm{R}}{\left[\mathrm{R}^{2}+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^{2}\right]}$
AIPMT- 2009
Alternating Current
155363
A resistance ' $R$ ' draws power ' $P$ ' when connected to an $\mathrm{AC}$ source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes ' $Z$ ' the power drawn will be
A Given that, Resistance $=\mathrm{R}$ Power $=\mathrm{P}$ We know that, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ When inductance is placed in series with resistance, Then, $\quad \mathrm{P}^{\prime}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\mathrm{rms}} \cos \phi$ In case of series $\mathrm{R}-\mathrm{L}$ circuit $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}$ $\mathrm{P}^{\prime}=\mathrm{V} \times\left(\frac{\mathrm{V}}{\mathrm{Z}}\right) \cdot \frac{\mathrm{R}}{\mathrm{Z}}$ $\mathrm{P}^{\prime}=\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}}$ Dividing equation (ii) by equation (i), we get- $\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}}}{\frac{\mathrm{V}^{2}}{\mathrm{R}}}=\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}} \times \frac{\mathrm{R}}{\mathrm{V}^{2}}$ $\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\mathrm{R}^{2}}{\mathrm{Z}^{2}}$ $\mathrm{P}^{\prime}=\mathrm{P}\left(\frac{\mathrm{R}}{\mathrm{Z}}\right)^{2}$
AIPMT- 2015
Alternating Current
155364
An inductor $20 \mathrm{mH}$, a capacitor $50 \mu \mathrm{F}$ and a resistor $40 \Omega$ are connected in series across a source of emf $V=10 \sin 340 t$. The power loss in $\mathrm{AC}$ circuit is
155361
A coil of inductive reactance $31 \Omega$ has a resistance of $8 \Omega$. it is placed in series with a condenser of capacitive reactance $25 \Omega$. The combination is connected to an $\mathrm{AC}$ source of $110 \mathrm{~V}$. The power factor of the circuit is
1 0.56
2 0.64
3 0.80
4 0.33
Explanation:
C Given that, $\mathrm{X}_{\mathrm{L}}=31 \Omega, \mathrm{R}=8 \Omega \&, \mathrm{X}_{\mathrm{C}}=25 \Omega$ According to question, We know that, impedance of circuit - $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{8^{2}+(31-25)^{2}}$ $Z=\sqrt{64+36}$ $Z=10$ $\therefore$ Power factor, $\cos \phi=\frac{R}{Z}=\frac{8}{10}$ $\operatorname{Cos} \phi=0.8$
AIPMT- 2006
Alternating Current
155362
Power dissipated in an L-C-R series circuit connected to an $\mathrm{AC}$ source of emf $\varepsilon$ is
A According to question, Impedance of L-R-C series circuit $|Z|=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $|Z|=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$ $\cos \phi=\frac{R}{|Z|}$ Power dissipated, $\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\mathrm{P}=\varepsilon \times \frac{\varepsilon}{|\mathrm{Z}|} \times \frac{\mathrm{R}}{|\mathrm{Z}|}$ $\mathrm{P}=\frac{\varepsilon^{2} \mathrm{R}}{|\mathrm{Z}|^{2}}=\frac{\varepsilon^{2} \mathrm{R}}{\left(\sqrt{\mathrm{R}^{2}+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^{2}}\right)^{2}}$ $\mathrm{P}=\frac{\varepsilon^{2} \mathrm{R}}{\left[\mathrm{R}^{2}+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^{2}\right]}$
AIPMT- 2009
Alternating Current
155363
A resistance ' $R$ ' draws power ' $P$ ' when connected to an $\mathrm{AC}$ source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes ' $Z$ ' the power drawn will be
A Given that, Resistance $=\mathrm{R}$ Power $=\mathrm{P}$ We know that, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ When inductance is placed in series with resistance, Then, $\quad \mathrm{P}^{\prime}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\mathrm{rms}} \cos \phi$ In case of series $\mathrm{R}-\mathrm{L}$ circuit $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}$ $\mathrm{P}^{\prime}=\mathrm{V} \times\left(\frac{\mathrm{V}}{\mathrm{Z}}\right) \cdot \frac{\mathrm{R}}{\mathrm{Z}}$ $\mathrm{P}^{\prime}=\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}}$ Dividing equation (ii) by equation (i), we get- $\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}}}{\frac{\mathrm{V}^{2}}{\mathrm{R}}}=\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}} \times \frac{\mathrm{R}}{\mathrm{V}^{2}}$ $\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\mathrm{R}^{2}}{\mathrm{Z}^{2}}$ $\mathrm{P}^{\prime}=\mathrm{P}\left(\frac{\mathrm{R}}{\mathrm{Z}}\right)^{2}$
AIPMT- 2015
Alternating Current
155364
An inductor $20 \mathrm{mH}$, a capacitor $50 \mu \mathrm{F}$ and a resistor $40 \Omega$ are connected in series across a source of emf $V=10 \sin 340 t$. The power loss in $\mathrm{AC}$ circuit is
155361
A coil of inductive reactance $31 \Omega$ has a resistance of $8 \Omega$. it is placed in series with a condenser of capacitive reactance $25 \Omega$. The combination is connected to an $\mathrm{AC}$ source of $110 \mathrm{~V}$. The power factor of the circuit is
1 0.56
2 0.64
3 0.80
4 0.33
Explanation:
C Given that, $\mathrm{X}_{\mathrm{L}}=31 \Omega, \mathrm{R}=8 \Omega \&, \mathrm{X}_{\mathrm{C}}=25 \Omega$ According to question, We know that, impedance of circuit - $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z=\sqrt{8^{2}+(31-25)^{2}}$ $Z=\sqrt{64+36}$ $Z=10$ $\therefore$ Power factor, $\cos \phi=\frac{R}{Z}=\frac{8}{10}$ $\operatorname{Cos} \phi=0.8$
AIPMT- 2006
Alternating Current
155362
Power dissipated in an L-C-R series circuit connected to an $\mathrm{AC}$ source of emf $\varepsilon$ is
A According to question, Impedance of L-R-C series circuit $|Z|=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $|Z|=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$ $\cos \phi=\frac{R}{|Z|}$ Power dissipated, $\mathrm{P}=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \phi$ $\mathrm{P}=\varepsilon \times \frac{\varepsilon}{|\mathrm{Z}|} \times \frac{\mathrm{R}}{|\mathrm{Z}|}$ $\mathrm{P}=\frac{\varepsilon^{2} \mathrm{R}}{|\mathrm{Z}|^{2}}=\frac{\varepsilon^{2} \mathrm{R}}{\left(\sqrt{\mathrm{R}^{2}+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^{2}}\right)^{2}}$ $\mathrm{P}=\frac{\varepsilon^{2} \mathrm{R}}{\left[\mathrm{R}^{2}+\left(\omega \mathrm{L}-\frac{1}{\omega \mathrm{C}}\right)^{2}\right]}$
AIPMT- 2009
Alternating Current
155363
A resistance ' $R$ ' draws power ' $P$ ' when connected to an $\mathrm{AC}$ source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes ' $Z$ ' the power drawn will be
A Given that, Resistance $=\mathrm{R}$ Power $=\mathrm{P}$ We know that, $\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}$ When inductance is placed in series with resistance, Then, $\quad \mathrm{P}^{\prime}=\mathrm{V}_{\text {rms }} \mathrm{I}_{\mathrm{rms}} \cos \phi$ In case of series $\mathrm{R}-\mathrm{L}$ circuit $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}$ $\mathrm{P}^{\prime}=\mathrm{V} \times\left(\frac{\mathrm{V}}{\mathrm{Z}}\right) \cdot \frac{\mathrm{R}}{\mathrm{Z}}$ $\mathrm{P}^{\prime}=\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}}$ Dividing equation (ii) by equation (i), we get- $\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}}}{\frac{\mathrm{V}^{2}}{\mathrm{R}}}=\frac{\mathrm{V}^{2} \mathrm{R}}{\mathrm{Z}^{2}} \times \frac{\mathrm{R}}{\mathrm{V}^{2}}$ $\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\mathrm{R}^{2}}{\mathrm{Z}^{2}}$ $\mathrm{P}^{\prime}=\mathrm{P}\left(\frac{\mathrm{R}}{\mathrm{Z}}\right)^{2}$
AIPMT- 2015
Alternating Current
155364
An inductor $20 \mathrm{mH}$, a capacitor $50 \mu \mathrm{F}$ and a resistor $40 \Omega$ are connected in series across a source of emf $V=10 \sin 340 t$. The power loss in $\mathrm{AC}$ circuit is
