356023
A \(60{\rm{ }}W/120{\rm{ }}V\) bulb is connected to a \(240V/60{\rm{ }}Hz\) supply with an inductance in series. Find the value of inductance so that bulb gets correct voltage.
356024
The emf \({E=4 \cos 1000 t}\) volts is applied to an \({L R}\) circuit, containing inductance of \(3\,mH\) and resistance of \({4 \Omega}\). The amplitude of current is
356025
A coil of inductive reactance \(1{\rm{/}}\sqrt 3 \Omega \) and resistance \(1\Omega \) is connected to a 200 \(V\), 50 \(Hz\) \(A.C\) supply. The time lag between maximum voltage and current is
1 \(\frac{1}{{600}}s\)
2 \(\frac{1}{{200}}s\)
3 \(\frac{1}{{300}}s\)
4 \(\frac{1}{{500}}s\)
Explanation:
Given,\({X_L} = \frac{1}{{\sqrt 3 }}\Omega ,R = 1\Omega ,V = 200V,v = 50Hz\) In series \(LR\) circuit, phase lag \(\phi \) is given by \(\tan \phi = \frac{{{X_L}}}{R} = \frac{{\frac{1}{{\sqrt 3 }}}}{1} = \frac{1}{{\sqrt 3 }} \Rightarrow \phi = \frac{\pi }{6}\) \(\therefore \,2\pi v\Delta t = \frac{\pi }{6},\) (where, \(\Delta t\) is time lag) \( \Rightarrow \Delta t = \frac{\pi }{{6 \times 2 \times \pi \times 50}} = \frac{1}{{600}}s\)
KCET - 2017
PHXII07:ALTERNATING CURRENT
356026
If a circuit made up of a resistance \(1\Omega \) and inductance 0.01 \(H\), and alternating emf 200\(V\) at 50 \(Hz\) is connected, then the phase difference between the current and the emf in the circuit is:
1 \({\tan ^{ - 1}}(\pi )\)
2 \({\tan ^{ - 1}}(\frac{\pi }{2})\)
3 \({\tan ^{ - 1}}(\frac{\pi }{4})\)
4 \({\tan ^{ - 1}}(\frac{\pi }{3})\)
Explanation:
The phase difference, \(\tan \phi = \frac{{{X_L}}}{R}\) and \({X_L} = \omega L = 2\pi fL = 2\pi \times 50 \times 0.01 = \pi \Omega \) Also \(R = 1\Omega \Rightarrow \phi = {\tan ^{ - 1}}(\pi )\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII07:ALTERNATING CURRENT
356023
A \(60{\rm{ }}W/120{\rm{ }}V\) bulb is connected to a \(240V/60{\rm{ }}Hz\) supply with an inductance in series. Find the value of inductance so that bulb gets correct voltage.
356024
The emf \({E=4 \cos 1000 t}\) volts is applied to an \({L R}\) circuit, containing inductance of \(3\,mH\) and resistance of \({4 \Omega}\). The amplitude of current is
356025
A coil of inductive reactance \(1{\rm{/}}\sqrt 3 \Omega \) and resistance \(1\Omega \) is connected to a 200 \(V\), 50 \(Hz\) \(A.C\) supply. The time lag between maximum voltage and current is
1 \(\frac{1}{{600}}s\)
2 \(\frac{1}{{200}}s\)
3 \(\frac{1}{{300}}s\)
4 \(\frac{1}{{500}}s\)
Explanation:
Given,\({X_L} = \frac{1}{{\sqrt 3 }}\Omega ,R = 1\Omega ,V = 200V,v = 50Hz\) In series \(LR\) circuit, phase lag \(\phi \) is given by \(\tan \phi = \frac{{{X_L}}}{R} = \frac{{\frac{1}{{\sqrt 3 }}}}{1} = \frac{1}{{\sqrt 3 }} \Rightarrow \phi = \frac{\pi }{6}\) \(\therefore \,2\pi v\Delta t = \frac{\pi }{6},\) (where, \(\Delta t\) is time lag) \( \Rightarrow \Delta t = \frac{\pi }{{6 \times 2 \times \pi \times 50}} = \frac{1}{{600}}s\)
KCET - 2017
PHXII07:ALTERNATING CURRENT
356026
If a circuit made up of a resistance \(1\Omega \) and inductance 0.01 \(H\), and alternating emf 200\(V\) at 50 \(Hz\) is connected, then the phase difference between the current and the emf in the circuit is:
1 \({\tan ^{ - 1}}(\pi )\)
2 \({\tan ^{ - 1}}(\frac{\pi }{2})\)
3 \({\tan ^{ - 1}}(\frac{\pi }{4})\)
4 \({\tan ^{ - 1}}(\frac{\pi }{3})\)
Explanation:
The phase difference, \(\tan \phi = \frac{{{X_L}}}{R}\) and \({X_L} = \omega L = 2\pi fL = 2\pi \times 50 \times 0.01 = \pi \Omega \) Also \(R = 1\Omega \Rightarrow \phi = {\tan ^{ - 1}}(\pi )\)
356023
A \(60{\rm{ }}W/120{\rm{ }}V\) bulb is connected to a \(240V/60{\rm{ }}Hz\) supply with an inductance in series. Find the value of inductance so that bulb gets correct voltage.
356024
The emf \({E=4 \cos 1000 t}\) volts is applied to an \({L R}\) circuit, containing inductance of \(3\,mH\) and resistance of \({4 \Omega}\). The amplitude of current is
356025
A coil of inductive reactance \(1{\rm{/}}\sqrt 3 \Omega \) and resistance \(1\Omega \) is connected to a 200 \(V\), 50 \(Hz\) \(A.C\) supply. The time lag between maximum voltage and current is
1 \(\frac{1}{{600}}s\)
2 \(\frac{1}{{200}}s\)
3 \(\frac{1}{{300}}s\)
4 \(\frac{1}{{500}}s\)
Explanation:
Given,\({X_L} = \frac{1}{{\sqrt 3 }}\Omega ,R = 1\Omega ,V = 200V,v = 50Hz\) In series \(LR\) circuit, phase lag \(\phi \) is given by \(\tan \phi = \frac{{{X_L}}}{R} = \frac{{\frac{1}{{\sqrt 3 }}}}{1} = \frac{1}{{\sqrt 3 }} \Rightarrow \phi = \frac{\pi }{6}\) \(\therefore \,2\pi v\Delta t = \frac{\pi }{6},\) (where, \(\Delta t\) is time lag) \( \Rightarrow \Delta t = \frac{\pi }{{6 \times 2 \times \pi \times 50}} = \frac{1}{{600}}s\)
KCET - 2017
PHXII07:ALTERNATING CURRENT
356026
If a circuit made up of a resistance \(1\Omega \) and inductance 0.01 \(H\), and alternating emf 200\(V\) at 50 \(Hz\) is connected, then the phase difference between the current and the emf in the circuit is:
1 \({\tan ^{ - 1}}(\pi )\)
2 \({\tan ^{ - 1}}(\frac{\pi }{2})\)
3 \({\tan ^{ - 1}}(\frac{\pi }{4})\)
4 \({\tan ^{ - 1}}(\frac{\pi }{3})\)
Explanation:
The phase difference, \(\tan \phi = \frac{{{X_L}}}{R}\) and \({X_L} = \omega L = 2\pi fL = 2\pi \times 50 \times 0.01 = \pi \Omega \) Also \(R = 1\Omega \Rightarrow \phi = {\tan ^{ - 1}}(\pi )\)
356023
A \(60{\rm{ }}W/120{\rm{ }}V\) bulb is connected to a \(240V/60{\rm{ }}Hz\) supply with an inductance in series. Find the value of inductance so that bulb gets correct voltage.
356024
The emf \({E=4 \cos 1000 t}\) volts is applied to an \({L R}\) circuit, containing inductance of \(3\,mH\) and resistance of \({4 \Omega}\). The amplitude of current is
356025
A coil of inductive reactance \(1{\rm{/}}\sqrt 3 \Omega \) and resistance \(1\Omega \) is connected to a 200 \(V\), 50 \(Hz\) \(A.C\) supply. The time lag between maximum voltage and current is
1 \(\frac{1}{{600}}s\)
2 \(\frac{1}{{200}}s\)
3 \(\frac{1}{{300}}s\)
4 \(\frac{1}{{500}}s\)
Explanation:
Given,\({X_L} = \frac{1}{{\sqrt 3 }}\Omega ,R = 1\Omega ,V = 200V,v = 50Hz\) In series \(LR\) circuit, phase lag \(\phi \) is given by \(\tan \phi = \frac{{{X_L}}}{R} = \frac{{\frac{1}{{\sqrt 3 }}}}{1} = \frac{1}{{\sqrt 3 }} \Rightarrow \phi = \frac{\pi }{6}\) \(\therefore \,2\pi v\Delta t = \frac{\pi }{6},\) (where, \(\Delta t\) is time lag) \( \Rightarrow \Delta t = \frac{\pi }{{6 \times 2 \times \pi \times 50}} = \frac{1}{{600}}s\)
KCET - 2017
PHXII07:ALTERNATING CURRENT
356026
If a circuit made up of a resistance \(1\Omega \) and inductance 0.01 \(H\), and alternating emf 200\(V\) at 50 \(Hz\) is connected, then the phase difference between the current and the emf in the circuit is:
1 \({\tan ^{ - 1}}(\pi )\)
2 \({\tan ^{ - 1}}(\frac{\pi }{2})\)
3 \({\tan ^{ - 1}}(\frac{\pi }{4})\)
4 \({\tan ^{ - 1}}(\frac{\pi }{3})\)
Explanation:
The phase difference, \(\tan \phi = \frac{{{X_L}}}{R}\) and \({X_L} = \omega L = 2\pi fL = 2\pi \times 50 \times 0.01 = \pi \Omega \) Also \(R = 1\Omega \Rightarrow \phi = {\tan ^{ - 1}}(\pi )\)