NEET Test Series from KOTA - 10 Papers In MS WORD
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Alternating Current
155186
In an LCR circuit having $L=8 \mathrm{H}, \mathrm{C}=0.5 \mu \mathrm{F}$ and $R=100 \Omega$ in series, the resonance frequency in $\mathrm{rad} / \mathrm{s}$ is
155187
The power factor of $L-C-R$ circuit at resonance is
1 0
2 $\frac{1}{2}$
3 $\frac{1}{\sqrt{2}}$
4 1
5 -1
Explanation:
D Let us consider on series R-L-C circuit shown in the figure given below - We know that, Impedance of LCR circuit, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ Power factor can be given by - $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$ At resonance condition- $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Then, $\quad \cos \phi=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}}$ $=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+0}} \quad \quad \text { (From eq }{ }^{\mathrm{n}} \text { (i)) }$ $=\frac{\mathrm{R}}{\mathrm{R}}=1$ So, $\quad \cos \phi=1$
KERALA CEE 2012
Alternating Current
155188
In an $L-C-R$ series circuit, the values of $R, X_{L}$ and $X_{C}$ are $120 \Omega, 180 \Omega$ and $130 \Omega$, what is the impedance of the circuit?
1 $120 \Omega$
2 $130 \Omega$
3 $180 \Omega$
4 $330 \Omega$
Explanation:
B Given, Resistance, $(\mathrm{R})=120 \Omega$, Inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=180 \Omega$, Capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)=130 \Omega$ Circuit impedance $(\mathrm{Z})=$ ? We know that, Impedance of LCR circuit, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z =\sqrt{(120)^{2}+(180-130)^{2}}$ $Z=\sqrt{14400+2500}, Z=\sqrt{16900}$ $\text { So, } \quad Z =130 \Omega$
Manipal UGET-2016
Alternating Current
155189
In an L-C -R circuit which one of the following statements is correct?
1 L and R oppose each other.
2 $\mathrm{R}$ value increase with frequency
3 Then inductive reactance increases with frequency
4 The capacitive reactance increases with frequency
Explanation:
C In an L-C-R circuit - Inductive reactance $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$ $X_{L}=2 \pi f L$ $X_{L} \propto f$ From the above expression we can see that value of inductive reactance directly proportional to frequency therefore inductive reactance increases with increase frequency. - $\quad \mathrm{L}$ and $\mathrm{C}$ oppose each other - $\mathrm{R}$ value constant with frequency - $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}} \Rightarrow \mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}}$
155186
In an LCR circuit having $L=8 \mathrm{H}, \mathrm{C}=0.5 \mu \mathrm{F}$ and $R=100 \Omega$ in series, the resonance frequency in $\mathrm{rad} / \mathrm{s}$ is
155187
The power factor of $L-C-R$ circuit at resonance is
1 0
2 $\frac{1}{2}$
3 $\frac{1}{\sqrt{2}}$
4 1
5 -1
Explanation:
D Let us consider on series R-L-C circuit shown in the figure given below - We know that, Impedance of LCR circuit, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ Power factor can be given by - $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$ At resonance condition- $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Then, $\quad \cos \phi=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}}$ $=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+0}} \quad \quad \text { (From eq }{ }^{\mathrm{n}} \text { (i)) }$ $=\frac{\mathrm{R}}{\mathrm{R}}=1$ So, $\quad \cos \phi=1$
KERALA CEE 2012
Alternating Current
155188
In an $L-C-R$ series circuit, the values of $R, X_{L}$ and $X_{C}$ are $120 \Omega, 180 \Omega$ and $130 \Omega$, what is the impedance of the circuit?
1 $120 \Omega$
2 $130 \Omega$
3 $180 \Omega$
4 $330 \Omega$
Explanation:
B Given, Resistance, $(\mathrm{R})=120 \Omega$, Inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=180 \Omega$, Capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)=130 \Omega$ Circuit impedance $(\mathrm{Z})=$ ? We know that, Impedance of LCR circuit, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z =\sqrt{(120)^{2}+(180-130)^{2}}$ $Z=\sqrt{14400+2500}, Z=\sqrt{16900}$ $\text { So, } \quad Z =130 \Omega$
Manipal UGET-2016
Alternating Current
155189
In an L-C -R circuit which one of the following statements is correct?
1 L and R oppose each other.
2 $\mathrm{R}$ value increase with frequency
3 Then inductive reactance increases with frequency
4 The capacitive reactance increases with frequency
Explanation:
C In an L-C-R circuit - Inductive reactance $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$ $X_{L}=2 \pi f L$ $X_{L} \propto f$ From the above expression we can see that value of inductive reactance directly proportional to frequency therefore inductive reactance increases with increase frequency. - $\quad \mathrm{L}$ and $\mathrm{C}$ oppose each other - $\mathrm{R}$ value constant with frequency - $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}} \Rightarrow \mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}}$
155186
In an LCR circuit having $L=8 \mathrm{H}, \mathrm{C}=0.5 \mu \mathrm{F}$ and $R=100 \Omega$ in series, the resonance frequency in $\mathrm{rad} / \mathrm{s}$ is
155187
The power factor of $L-C-R$ circuit at resonance is
1 0
2 $\frac{1}{2}$
3 $\frac{1}{\sqrt{2}}$
4 1
5 -1
Explanation:
D Let us consider on series R-L-C circuit shown in the figure given below - We know that, Impedance of LCR circuit, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ Power factor can be given by - $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$ At resonance condition- $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Then, $\quad \cos \phi=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}}$ $=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+0}} \quad \quad \text { (From eq }{ }^{\mathrm{n}} \text { (i)) }$ $=\frac{\mathrm{R}}{\mathrm{R}}=1$ So, $\quad \cos \phi=1$
KERALA CEE 2012
Alternating Current
155188
In an $L-C-R$ series circuit, the values of $R, X_{L}$ and $X_{C}$ are $120 \Omega, 180 \Omega$ and $130 \Omega$, what is the impedance of the circuit?
1 $120 \Omega$
2 $130 \Omega$
3 $180 \Omega$
4 $330 \Omega$
Explanation:
B Given, Resistance, $(\mathrm{R})=120 \Omega$, Inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=180 \Omega$, Capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)=130 \Omega$ Circuit impedance $(\mathrm{Z})=$ ? We know that, Impedance of LCR circuit, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z =\sqrt{(120)^{2}+(180-130)^{2}}$ $Z=\sqrt{14400+2500}, Z=\sqrt{16900}$ $\text { So, } \quad Z =130 \Omega$
Manipal UGET-2016
Alternating Current
155189
In an L-C -R circuit which one of the following statements is correct?
1 L and R oppose each other.
2 $\mathrm{R}$ value increase with frequency
3 Then inductive reactance increases with frequency
4 The capacitive reactance increases with frequency
Explanation:
C In an L-C-R circuit - Inductive reactance $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$ $X_{L}=2 \pi f L$ $X_{L} \propto f$ From the above expression we can see that value of inductive reactance directly proportional to frequency therefore inductive reactance increases with increase frequency. - $\quad \mathrm{L}$ and $\mathrm{C}$ oppose each other - $\mathrm{R}$ value constant with frequency - $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}} \Rightarrow \mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}}$
155186
In an LCR circuit having $L=8 \mathrm{H}, \mathrm{C}=0.5 \mu \mathrm{F}$ and $R=100 \Omega$ in series, the resonance frequency in $\mathrm{rad} / \mathrm{s}$ is
155187
The power factor of $L-C-R$ circuit at resonance is
1 0
2 $\frac{1}{2}$
3 $\frac{1}{\sqrt{2}}$
4 1
5 -1
Explanation:
D Let us consider on series R-L-C circuit shown in the figure given below - We know that, Impedance of LCR circuit, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ Power factor can be given by - $\cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$ At resonance condition- $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ Then, $\quad \cos \phi=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}}$ $=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+0}} \quad \quad \text { (From eq }{ }^{\mathrm{n}} \text { (i)) }$ $=\frac{\mathrm{R}}{\mathrm{R}}=1$ So, $\quad \cos \phi=1$
KERALA CEE 2012
Alternating Current
155188
In an $L-C-R$ series circuit, the values of $R, X_{L}$ and $X_{C}$ are $120 \Omega, 180 \Omega$ and $130 \Omega$, what is the impedance of the circuit?
1 $120 \Omega$
2 $130 \Omega$
3 $180 \Omega$
4 $330 \Omega$
Explanation:
B Given, Resistance, $(\mathrm{R})=120 \Omega$, Inductive reactance $\left(\mathrm{X}_{\mathrm{L}}\right)=180 \Omega$, Capacitive reactance $\left(\mathrm{X}_{\mathrm{C}}\right)=130 \Omega$ Circuit impedance $(\mathrm{Z})=$ ? We know that, Impedance of LCR circuit, $Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$ $Z =\sqrt{(120)^{2}+(180-130)^{2}}$ $Z=\sqrt{14400+2500}, Z=\sqrt{16900}$ $\text { So, } \quad Z =130 \Omega$
Manipal UGET-2016
Alternating Current
155189
In an L-C -R circuit which one of the following statements is correct?
1 L and R oppose each other.
2 $\mathrm{R}$ value increase with frequency
3 Then inductive reactance increases with frequency
4 The capacitive reactance increases with frequency
Explanation:
C In an L-C-R circuit - Inductive reactance $\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}$ $X_{L}=2 \pi f L$ $X_{L} \propto f$ From the above expression we can see that value of inductive reactance directly proportional to frequency therefore inductive reactance increases with increase frequency. - $\quad \mathrm{L}$ and $\mathrm{C}$ oppose each other - $\mathrm{R}$ value constant with frequency - $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}} \Rightarrow \mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}}$
