NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155355
The power factor of an A.C. circuit having resistance $R$ and inductance $L$ connected in series to an A.C. source of angular frequency $\omega$ is
C We know that, inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=\omega \mathrm{L}$ Impedance $(Z)=\sqrt{R^{2}+X_{L}^{2}}$ $Z=\sqrt{R^{2}+(\omega L)^{2}}$ $Z=\sqrt{R^{2}+\omega^{2} L^{2}}$ Power factor of AC circuit, $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}$
UP CPMT-2013
Alternating Current
155356
The power factor of a series L-C-R circuit when at resonance is
1 zero
2 0.5
3 1.0
4 depends on values of L,C and R
Explanation:
C In series R, L, C circuit at the condition of resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ We know that, $\therefore$ Impedance $(\mathrm{Z})=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$ $\mathrm{Z}=\mathrm{R}$ $\therefore$ Power factor, $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{R}}=1$ $\cos \phi=1$
UP CPMT-2005
Alternating Current
155358
In an $A C$ circuit $V$ and $I$ are given below, then find the power dissipated in the circuit $V=\mathbf{5 0}$ $\sin (50 t)$ volt and $I=50 \sin \left(50 t+\frac{\pi}{3}\right) \mathrm{mA}$
155359
In an $\mathrm{AC}$ circuit the emf (V) and the current (i) at any instant are given respectively by $V=V_{0} \sin \omega t, i=i_{0} \sin (\omega t-\phi)$ The average power in the circuit over one cycle of $\mathrm{AC}$ is
155355
The power factor of an A.C. circuit having resistance $R$ and inductance $L$ connected in series to an A.C. source of angular frequency $\omega$ is
C We know that, inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=\omega \mathrm{L}$ Impedance $(Z)=\sqrt{R^{2}+X_{L}^{2}}$ $Z=\sqrt{R^{2}+(\omega L)^{2}}$ $Z=\sqrt{R^{2}+\omega^{2} L^{2}}$ Power factor of AC circuit, $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}$
UP CPMT-2013
Alternating Current
155356
The power factor of a series L-C-R circuit when at resonance is
1 zero
2 0.5
3 1.0
4 depends on values of L,C and R
Explanation:
C In series R, L, C circuit at the condition of resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ We know that, $\therefore$ Impedance $(\mathrm{Z})=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$ $\mathrm{Z}=\mathrm{R}$ $\therefore$ Power factor, $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{R}}=1$ $\cos \phi=1$
UP CPMT-2005
Alternating Current
155358
In an $A C$ circuit $V$ and $I$ are given below, then find the power dissipated in the circuit $V=\mathbf{5 0}$ $\sin (50 t)$ volt and $I=50 \sin \left(50 t+\frac{\pi}{3}\right) \mathrm{mA}$
155359
In an $\mathrm{AC}$ circuit the emf (V) and the current (i) at any instant are given respectively by $V=V_{0} \sin \omega t, i=i_{0} \sin (\omega t-\phi)$ The average power in the circuit over one cycle of $\mathrm{AC}$ is
155355
The power factor of an A.C. circuit having resistance $R$ and inductance $L$ connected in series to an A.C. source of angular frequency $\omega$ is
C We know that, inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=\omega \mathrm{L}$ Impedance $(Z)=\sqrt{R^{2}+X_{L}^{2}}$ $Z=\sqrt{R^{2}+(\omega L)^{2}}$ $Z=\sqrt{R^{2}+\omega^{2} L^{2}}$ Power factor of AC circuit, $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}$
UP CPMT-2013
Alternating Current
155356
The power factor of a series L-C-R circuit when at resonance is
1 zero
2 0.5
3 1.0
4 depends on values of L,C and R
Explanation:
C In series R, L, C circuit at the condition of resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ We know that, $\therefore$ Impedance $(\mathrm{Z})=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$ $\mathrm{Z}=\mathrm{R}$ $\therefore$ Power factor, $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{R}}=1$ $\cos \phi=1$
UP CPMT-2005
Alternating Current
155358
In an $A C$ circuit $V$ and $I$ are given below, then find the power dissipated in the circuit $V=\mathbf{5 0}$ $\sin (50 t)$ volt and $I=50 \sin \left(50 t+\frac{\pi}{3}\right) \mathrm{mA}$
155359
In an $\mathrm{AC}$ circuit the emf (V) and the current (i) at any instant are given respectively by $V=V_{0} \sin \omega t, i=i_{0} \sin (\omega t-\phi)$ The average power in the circuit over one cycle of $\mathrm{AC}$ is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Alternating Current
155355
The power factor of an A.C. circuit having resistance $R$ and inductance $L$ connected in series to an A.C. source of angular frequency $\omega$ is
C We know that, inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=\omega \mathrm{L}$ Impedance $(Z)=\sqrt{R^{2}+X_{L}^{2}}$ $Z=\sqrt{R^{2}+(\omega L)^{2}}$ $Z=\sqrt{R^{2}+\omega^{2} L^{2}}$ Power factor of AC circuit, $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}$
UP CPMT-2013
Alternating Current
155356
The power factor of a series L-C-R circuit when at resonance is
1 zero
2 0.5
3 1.0
4 depends on values of L,C and R
Explanation:
C In series R, L, C circuit at the condition of resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ We know that, $\therefore$ Impedance $(\mathrm{Z})=\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}$ $\mathrm{Z}=\mathrm{R}$ $\therefore$ Power factor, $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{R}}{\mathrm{R}}=1$ $\cos \phi=1$
UP CPMT-2005
Alternating Current
155358
In an $A C$ circuit $V$ and $I$ are given below, then find the power dissipated in the circuit $V=\mathbf{5 0}$ $\sin (50 t)$ volt and $I=50 \sin \left(50 t+\frac{\pi}{3}\right) \mathrm{mA}$
155359
In an $\mathrm{AC}$ circuit the emf (V) and the current (i) at any instant are given respectively by $V=V_{0} \sin \omega t, i=i_{0} \sin (\omega t-\phi)$ The average power in the circuit over one cycle of $\mathrm{AC}$ is
