NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155259
IN an L-C-R circuit, if impedance is $\sqrt{3}$ times of resistance and capacitive reactance is zero. Find the phase difference.
1 Zero
2 $30^{\circ}$
3 $60^{\circ}$
4 data is incomplete
Explanation:
C We know that, $\tan \phi=\frac{\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}$ $\mathrm{X}_{\mathrm{C}}=0, \mathrm{X}_{\mathrm{L}}=\sqrt{3} \mathrm{R}$ $\tan \phi=\frac{\sqrt{3} \mathrm{R}}{\mathrm{R}}$ $\tan \phi=\sqrt{3}$ $\phi=\tan ^{-1}(\sqrt{3})$ $\phi=60^{\circ}$
UP CPMT-2003
Alternating Current
155260
A resistor $R$, and inductor $L$ and a capacitor $C$ are connected in series to a source of frequency $n$. If the resonant frequency is $n_{r}$, then the current lags behind voltage when :
1 $\mathrm{n}=0$
2 n $ \lt \mathrm{n}_{\mathrm{r}}$
3 $\mathrm{n}=\mathrm{n}_{\mathrm{r}}$
4 $n>n_{r}$
Explanation:
D When the reactance of inductance is more than reactance of condenser then current lag behind the voltage - $\mathrm{X}_{\mathrm{L}} >\mathrm{X}_{\mathrm{C}}$ $\omega \mathrm{L} \lt \frac{1}{\omega \mathrm{C}}$ $\text { Or } \omega>\frac{1}{\sqrt{\mathrm{LC}}}$ Then, $\mathrm{n}>\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ Or $\quad n>n_{r}$ ( $\mathrm{n}_{\mathrm{r}}$ is resonant frequency)
SRMJEEE - 2007
Alternating Current
155261
For a series LCR circuit, the rms values of voltage across various components are $V_{L}=90 \mathrm{~V}, V_{C}=60 \mathrm{~V}$ and $V_{R}=40 \mathrm{~V}$ Volt, The rms value of the voltage applied to the circuit is :
1 $190 \mathrm{~V}$
2 $110 \mathrm{~V}$
3 $70 \mathrm{~V}$
4 $50 \mathrm{~V}$
Explanation:
D Given that, Voltage across inductor $\left(\mathrm{V}_{\mathrm{L}}\right)=90 \mathrm{~V}$ Voltage across capacitor $\left(\mathrm{V}_{\mathrm{C}}\right)=60 \mathrm{~V}$ Voltage across resistance $\left(\mathrm{V}_{\mathrm{R}}\right)=40 \mathrm{~V}$ $\text { Now, } \mathrm{V}_{\mathrm{rms}} =\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$ $\mathrm{~V}_{\mathrm{rms}} =\sqrt{(40)^{2}+(90-60)^{2}}$ $\mathrm{~V}_{\mathrm{rms}} =\sqrt{(40)^{2}+(30)^{2}}$ $\mathrm{~V}_{\mathrm{rms}} =50 \mathrm{~V}$ Hence, the rms value of the voltage applied to the circuit is $50 \mathrm{~V}$.
MP PMT-2013
Alternating Current
155262
In a series $L-C-R$ circuit, resistance $R=10 \Omega$ and the impedance $Z=10 \Omega$. The phase difference between the current and the voltage is
1 $0^{\circ}$
2 $30^{\circ}$
3 $45^{\circ}$
4 $60^{\circ}$
Explanation:
A Given that, Resistance $(\mathrm{R})=10 \Omega$ Impedance $(Z)=10 \Omega$ We know that, phase difference - $\cos \phi =\frac{\mathrm{R}}{\mathrm{Z}}=\frac{10}{10}$ $\cos \phi =1$ $\phi =\cos ^{-1}(1)$ $\phi =0^{\circ}$
155259
IN an L-C-R circuit, if impedance is $\sqrt{3}$ times of resistance and capacitive reactance is zero. Find the phase difference.
1 Zero
2 $30^{\circ}$
3 $60^{\circ}$
4 data is incomplete
Explanation:
C We know that, $\tan \phi=\frac{\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}$ $\mathrm{X}_{\mathrm{C}}=0, \mathrm{X}_{\mathrm{L}}=\sqrt{3} \mathrm{R}$ $\tan \phi=\frac{\sqrt{3} \mathrm{R}}{\mathrm{R}}$ $\tan \phi=\sqrt{3}$ $\phi=\tan ^{-1}(\sqrt{3})$ $\phi=60^{\circ}$
UP CPMT-2003
Alternating Current
155260
A resistor $R$, and inductor $L$ and a capacitor $C$ are connected in series to a source of frequency $n$. If the resonant frequency is $n_{r}$, then the current lags behind voltage when :
1 $\mathrm{n}=0$
2 n $ \lt \mathrm{n}_{\mathrm{r}}$
3 $\mathrm{n}=\mathrm{n}_{\mathrm{r}}$
4 $n>n_{r}$
Explanation:
D When the reactance of inductance is more than reactance of condenser then current lag behind the voltage - $\mathrm{X}_{\mathrm{L}} >\mathrm{X}_{\mathrm{C}}$ $\omega \mathrm{L} \lt \frac{1}{\omega \mathrm{C}}$ $\text { Or } \omega>\frac{1}{\sqrt{\mathrm{LC}}}$ Then, $\mathrm{n}>\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ Or $\quad n>n_{r}$ ( $\mathrm{n}_{\mathrm{r}}$ is resonant frequency)
SRMJEEE - 2007
Alternating Current
155261
For a series LCR circuit, the rms values of voltage across various components are $V_{L}=90 \mathrm{~V}, V_{C}=60 \mathrm{~V}$ and $V_{R}=40 \mathrm{~V}$ Volt, The rms value of the voltage applied to the circuit is :
1 $190 \mathrm{~V}$
2 $110 \mathrm{~V}$
3 $70 \mathrm{~V}$
4 $50 \mathrm{~V}$
Explanation:
D Given that, Voltage across inductor $\left(\mathrm{V}_{\mathrm{L}}\right)=90 \mathrm{~V}$ Voltage across capacitor $\left(\mathrm{V}_{\mathrm{C}}\right)=60 \mathrm{~V}$ Voltage across resistance $\left(\mathrm{V}_{\mathrm{R}}\right)=40 \mathrm{~V}$ $\text { Now, } \mathrm{V}_{\mathrm{rms}} =\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$ $\mathrm{~V}_{\mathrm{rms}} =\sqrt{(40)^{2}+(90-60)^{2}}$ $\mathrm{~V}_{\mathrm{rms}} =\sqrt{(40)^{2}+(30)^{2}}$ $\mathrm{~V}_{\mathrm{rms}} =50 \mathrm{~V}$ Hence, the rms value of the voltage applied to the circuit is $50 \mathrm{~V}$.
MP PMT-2013
Alternating Current
155262
In a series $L-C-R$ circuit, resistance $R=10 \Omega$ and the impedance $Z=10 \Omega$. The phase difference between the current and the voltage is
1 $0^{\circ}$
2 $30^{\circ}$
3 $45^{\circ}$
4 $60^{\circ}$
Explanation:
A Given that, Resistance $(\mathrm{R})=10 \Omega$ Impedance $(Z)=10 \Omega$ We know that, phase difference - $\cos \phi =\frac{\mathrm{R}}{\mathrm{Z}}=\frac{10}{10}$ $\cos \phi =1$ $\phi =\cos ^{-1}(1)$ $\phi =0^{\circ}$
155259
IN an L-C-R circuit, if impedance is $\sqrt{3}$ times of resistance and capacitive reactance is zero. Find the phase difference.
1 Zero
2 $30^{\circ}$
3 $60^{\circ}$
4 data is incomplete
Explanation:
C We know that, $\tan \phi=\frac{\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}$ $\mathrm{X}_{\mathrm{C}}=0, \mathrm{X}_{\mathrm{L}}=\sqrt{3} \mathrm{R}$ $\tan \phi=\frac{\sqrt{3} \mathrm{R}}{\mathrm{R}}$ $\tan \phi=\sqrt{3}$ $\phi=\tan ^{-1}(\sqrt{3})$ $\phi=60^{\circ}$
UP CPMT-2003
Alternating Current
155260
A resistor $R$, and inductor $L$ and a capacitor $C$ are connected in series to a source of frequency $n$. If the resonant frequency is $n_{r}$, then the current lags behind voltage when :
1 $\mathrm{n}=0$
2 n $ \lt \mathrm{n}_{\mathrm{r}}$
3 $\mathrm{n}=\mathrm{n}_{\mathrm{r}}$
4 $n>n_{r}$
Explanation:
D When the reactance of inductance is more than reactance of condenser then current lag behind the voltage - $\mathrm{X}_{\mathrm{L}} >\mathrm{X}_{\mathrm{C}}$ $\omega \mathrm{L} \lt \frac{1}{\omega \mathrm{C}}$ $\text { Or } \omega>\frac{1}{\sqrt{\mathrm{LC}}}$ Then, $\mathrm{n}>\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ Or $\quad n>n_{r}$ ( $\mathrm{n}_{\mathrm{r}}$ is resonant frequency)
SRMJEEE - 2007
Alternating Current
155261
For a series LCR circuit, the rms values of voltage across various components are $V_{L}=90 \mathrm{~V}, V_{C}=60 \mathrm{~V}$ and $V_{R}=40 \mathrm{~V}$ Volt, The rms value of the voltage applied to the circuit is :
1 $190 \mathrm{~V}$
2 $110 \mathrm{~V}$
3 $70 \mathrm{~V}$
4 $50 \mathrm{~V}$
Explanation:
D Given that, Voltage across inductor $\left(\mathrm{V}_{\mathrm{L}}\right)=90 \mathrm{~V}$ Voltage across capacitor $\left(\mathrm{V}_{\mathrm{C}}\right)=60 \mathrm{~V}$ Voltage across resistance $\left(\mathrm{V}_{\mathrm{R}}\right)=40 \mathrm{~V}$ $\text { Now, } \mathrm{V}_{\mathrm{rms}} =\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$ $\mathrm{~V}_{\mathrm{rms}} =\sqrt{(40)^{2}+(90-60)^{2}}$ $\mathrm{~V}_{\mathrm{rms}} =\sqrt{(40)^{2}+(30)^{2}}$ $\mathrm{~V}_{\mathrm{rms}} =50 \mathrm{~V}$ Hence, the rms value of the voltage applied to the circuit is $50 \mathrm{~V}$.
MP PMT-2013
Alternating Current
155262
In a series $L-C-R$ circuit, resistance $R=10 \Omega$ and the impedance $Z=10 \Omega$. The phase difference between the current and the voltage is
1 $0^{\circ}$
2 $30^{\circ}$
3 $45^{\circ}$
4 $60^{\circ}$
Explanation:
A Given that, Resistance $(\mathrm{R})=10 \Omega$ Impedance $(Z)=10 \Omega$ We know that, phase difference - $\cos \phi =\frac{\mathrm{R}}{\mathrm{Z}}=\frac{10}{10}$ $\cos \phi =1$ $\phi =\cos ^{-1}(1)$ $\phi =0^{\circ}$
155259
IN an L-C-R circuit, if impedance is $\sqrt{3}$ times of resistance and capacitive reactance is zero. Find the phase difference.
1 Zero
2 $30^{\circ}$
3 $60^{\circ}$
4 data is incomplete
Explanation:
C We know that, $\tan \phi=\frac{\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}$ $\mathrm{X}_{\mathrm{C}}=0, \mathrm{X}_{\mathrm{L}}=\sqrt{3} \mathrm{R}$ $\tan \phi=\frac{\sqrt{3} \mathrm{R}}{\mathrm{R}}$ $\tan \phi=\sqrt{3}$ $\phi=\tan ^{-1}(\sqrt{3})$ $\phi=60^{\circ}$
UP CPMT-2003
Alternating Current
155260
A resistor $R$, and inductor $L$ and a capacitor $C$ are connected in series to a source of frequency $n$. If the resonant frequency is $n_{r}$, then the current lags behind voltage when :
1 $\mathrm{n}=0$
2 n $ \lt \mathrm{n}_{\mathrm{r}}$
3 $\mathrm{n}=\mathrm{n}_{\mathrm{r}}$
4 $n>n_{r}$
Explanation:
D When the reactance of inductance is more than reactance of condenser then current lag behind the voltage - $\mathrm{X}_{\mathrm{L}} >\mathrm{X}_{\mathrm{C}}$ $\omega \mathrm{L} \lt \frac{1}{\omega \mathrm{C}}$ $\text { Or } \omega>\frac{1}{\sqrt{\mathrm{LC}}}$ Then, $\mathrm{n}>\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ Or $\quad n>n_{r}$ ( $\mathrm{n}_{\mathrm{r}}$ is resonant frequency)
SRMJEEE - 2007
Alternating Current
155261
For a series LCR circuit, the rms values of voltage across various components are $V_{L}=90 \mathrm{~V}, V_{C}=60 \mathrm{~V}$ and $V_{R}=40 \mathrm{~V}$ Volt, The rms value of the voltage applied to the circuit is :
1 $190 \mathrm{~V}$
2 $110 \mathrm{~V}$
3 $70 \mathrm{~V}$
4 $50 \mathrm{~V}$
Explanation:
D Given that, Voltage across inductor $\left(\mathrm{V}_{\mathrm{L}}\right)=90 \mathrm{~V}$ Voltage across capacitor $\left(\mathrm{V}_{\mathrm{C}}\right)=60 \mathrm{~V}$ Voltage across resistance $\left(\mathrm{V}_{\mathrm{R}}\right)=40 \mathrm{~V}$ $\text { Now, } \mathrm{V}_{\mathrm{rms}} =\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$ $\mathrm{~V}_{\mathrm{rms}} =\sqrt{(40)^{2}+(90-60)^{2}}$ $\mathrm{~V}_{\mathrm{rms}} =\sqrt{(40)^{2}+(30)^{2}}$ $\mathrm{~V}_{\mathrm{rms}} =50 \mathrm{~V}$ Hence, the rms value of the voltage applied to the circuit is $50 \mathrm{~V}$.
MP PMT-2013
Alternating Current
155262
In a series $L-C-R$ circuit, resistance $R=10 \Omega$ and the impedance $Z=10 \Omega$. The phase difference between the current and the voltage is
1 $0^{\circ}$
2 $30^{\circ}$
3 $45^{\circ}$
4 $60^{\circ}$
Explanation:
A Given that, Resistance $(\mathrm{R})=10 \Omega$ Impedance $(Z)=10 \Omega$ We know that, phase difference - $\cos \phi =\frac{\mathrm{R}}{\mathrm{Z}}=\frac{10}{10}$ $\cos \phi =1$ $\phi =\cos ^{-1}(1)$ $\phi =0^{\circ}$
