356175
The coefficient of induction of a choke coil is \(0.1\,H\) and resistance is \(12\,\Omega \). If it is connected to an a.c source of frequency \(60\;Hz\). Then the power factor will be:
1 0.35
2 0.30
3 0.28
4 0.24
Explanation:
As we know (with \(\omega=2 \pi f\) ) \(\cos \phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}\) \(=\dfrac{12}{\sqrt{(12)^{2}+4 \times \pi^{2} \times(60)^{2} \times(0.1)^{2}}}\) \(\Rightarrow \cos \phi=0.30\) So, correct option is (2).
PHXII07:ALTERNATING CURRENT
356176
In a series \({L R}\) circuit, the voltage drop across inductor is \(8\,V\) and across resistor is \(6\,V\) . The voltage applied and power factor of circuit respectively are
1 \({14 V, 0.8}\)
2 \({10 {~V}, 0.8}\)
3 \({10 {~V}, 0.6}\)
4 \({14 V, 0.6}\)
Explanation:
\({V_{L}=8 V, V_{R}=6 V}\), \({V=\sqrt{V_{L}^{2}+V_{R}^{2}}=10 V}\) Power factor, \({\cos \phi=\dfrac{V_{R}}{V}=\dfrac{6}{10}=0.6}\)
PHXII07:ALTERNATING CURRENT
356177
The power factor of an \(R\)-\(L\) circuit is \(\frac{1}{{\sqrt 2 }}\). If the frequency of \(AC\) is doubled, then what will be the power factor?
356178
An \(A\). \(C\) voltage \(E = 200\sin 300\,t\) is applied across a series combination of \(R = 10\,\Omega \) and \(L = 800\,mH\). Calculate the power factor of the circuit.
1 0.0315
2 0.416
3 0.0416
4 0.417
Explanation:
\(X_{L}=\omega L=300 \times 800 \times 10^{-3}=240 \Omega\) \(Z=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{(10)^{2}+(240)^{2}}=240.2 \Omega\) Power factor, \(\cos \phi=\dfrac{R}{Z}=\dfrac{10}{240.2}=0.0416\) So, correct option is (3).
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII07:ALTERNATING CURRENT
356175
The coefficient of induction of a choke coil is \(0.1\,H\) and resistance is \(12\,\Omega \). If it is connected to an a.c source of frequency \(60\;Hz\). Then the power factor will be:
1 0.35
2 0.30
3 0.28
4 0.24
Explanation:
As we know (with \(\omega=2 \pi f\) ) \(\cos \phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}\) \(=\dfrac{12}{\sqrt{(12)^{2}+4 \times \pi^{2} \times(60)^{2} \times(0.1)^{2}}}\) \(\Rightarrow \cos \phi=0.30\) So, correct option is (2).
PHXII07:ALTERNATING CURRENT
356176
In a series \({L R}\) circuit, the voltage drop across inductor is \(8\,V\) and across resistor is \(6\,V\) . The voltage applied and power factor of circuit respectively are
1 \({14 V, 0.8}\)
2 \({10 {~V}, 0.8}\)
3 \({10 {~V}, 0.6}\)
4 \({14 V, 0.6}\)
Explanation:
\({V_{L}=8 V, V_{R}=6 V}\), \({V=\sqrt{V_{L}^{2}+V_{R}^{2}}=10 V}\) Power factor, \({\cos \phi=\dfrac{V_{R}}{V}=\dfrac{6}{10}=0.6}\)
PHXII07:ALTERNATING CURRENT
356177
The power factor of an \(R\)-\(L\) circuit is \(\frac{1}{{\sqrt 2 }}\). If the frequency of \(AC\) is doubled, then what will be the power factor?
356178
An \(A\). \(C\) voltage \(E = 200\sin 300\,t\) is applied across a series combination of \(R = 10\,\Omega \) and \(L = 800\,mH\). Calculate the power factor of the circuit.
1 0.0315
2 0.416
3 0.0416
4 0.417
Explanation:
\(X_{L}=\omega L=300 \times 800 \times 10^{-3}=240 \Omega\) \(Z=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{(10)^{2}+(240)^{2}}=240.2 \Omega\) Power factor, \(\cos \phi=\dfrac{R}{Z}=\dfrac{10}{240.2}=0.0416\) So, correct option is (3).
356175
The coefficient of induction of a choke coil is \(0.1\,H\) and resistance is \(12\,\Omega \). If it is connected to an a.c source of frequency \(60\;Hz\). Then the power factor will be:
1 0.35
2 0.30
3 0.28
4 0.24
Explanation:
As we know (with \(\omega=2 \pi f\) ) \(\cos \phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}\) \(=\dfrac{12}{\sqrt{(12)^{2}+4 \times \pi^{2} \times(60)^{2} \times(0.1)^{2}}}\) \(\Rightarrow \cos \phi=0.30\) So, correct option is (2).
PHXII07:ALTERNATING CURRENT
356176
In a series \({L R}\) circuit, the voltage drop across inductor is \(8\,V\) and across resistor is \(6\,V\) . The voltage applied and power factor of circuit respectively are
1 \({14 V, 0.8}\)
2 \({10 {~V}, 0.8}\)
3 \({10 {~V}, 0.6}\)
4 \({14 V, 0.6}\)
Explanation:
\({V_{L}=8 V, V_{R}=6 V}\), \({V=\sqrt{V_{L}^{2}+V_{R}^{2}}=10 V}\) Power factor, \({\cos \phi=\dfrac{V_{R}}{V}=\dfrac{6}{10}=0.6}\)
PHXII07:ALTERNATING CURRENT
356177
The power factor of an \(R\)-\(L\) circuit is \(\frac{1}{{\sqrt 2 }}\). If the frequency of \(AC\) is doubled, then what will be the power factor?
356178
An \(A\). \(C\) voltage \(E = 200\sin 300\,t\) is applied across a series combination of \(R = 10\,\Omega \) and \(L = 800\,mH\). Calculate the power factor of the circuit.
1 0.0315
2 0.416
3 0.0416
4 0.417
Explanation:
\(X_{L}=\omega L=300 \times 800 \times 10^{-3}=240 \Omega\) \(Z=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{(10)^{2}+(240)^{2}}=240.2 \Omega\) Power factor, \(\cos \phi=\dfrac{R}{Z}=\dfrac{10}{240.2}=0.0416\) So, correct option is (3).
356175
The coefficient of induction of a choke coil is \(0.1\,H\) and resistance is \(12\,\Omega \). If it is connected to an a.c source of frequency \(60\;Hz\). Then the power factor will be:
1 0.35
2 0.30
3 0.28
4 0.24
Explanation:
As we know (with \(\omega=2 \pi f\) ) \(\cos \phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}\) \(=\dfrac{12}{\sqrt{(12)^{2}+4 \times \pi^{2} \times(60)^{2} \times(0.1)^{2}}}\) \(\Rightarrow \cos \phi=0.30\) So, correct option is (2).
PHXII07:ALTERNATING CURRENT
356176
In a series \({L R}\) circuit, the voltage drop across inductor is \(8\,V\) and across resistor is \(6\,V\) . The voltage applied and power factor of circuit respectively are
1 \({14 V, 0.8}\)
2 \({10 {~V}, 0.8}\)
3 \({10 {~V}, 0.6}\)
4 \({14 V, 0.6}\)
Explanation:
\({V_{L}=8 V, V_{R}=6 V}\), \({V=\sqrt{V_{L}^{2}+V_{R}^{2}}=10 V}\) Power factor, \({\cos \phi=\dfrac{V_{R}}{V}=\dfrac{6}{10}=0.6}\)
PHXII07:ALTERNATING CURRENT
356177
The power factor of an \(R\)-\(L\) circuit is \(\frac{1}{{\sqrt 2 }}\). If the frequency of \(AC\) is doubled, then what will be the power factor?
356178
An \(A\). \(C\) voltage \(E = 200\sin 300\,t\) is applied across a series combination of \(R = 10\,\Omega \) and \(L = 800\,mH\). Calculate the power factor of the circuit.
1 0.0315
2 0.416
3 0.0416
4 0.417
Explanation:
\(X_{L}=\omega L=300 \times 800 \times 10^{-3}=240 \Omega\) \(Z=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{(10)^{2}+(240)^{2}}=240.2 \Omega\) Power factor, \(\cos \phi=\dfrac{R}{Z}=\dfrac{10}{240.2}=0.0416\) So, correct option is (3).
