NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII07:ALTERNATING CURRENT
356050
The peak value of an alternating \(e.m.f.\) \({E}\) is given by \({E=E_{0} \cos \omega t}\) is \(10\,volts\) and its frequency is \(50\,Hz\). At time \({t=\dfrac{1}{600} s}\), the instantaneous emf is
356051
An \(ac\) voltage is represented by \(E = 220\sqrt 2 \cos (50\pi )t\) How many times will the current becomes zero in 1\(s\)?
1 \({\rm{100}}\,{\rm{times}}\)
2 \({\rm{50}}\,{\rm{times}}\)
3 \({\rm{25}}\,{\rm{times}}\)
4 \({\rm{30}}\,{\rm{times}}\)
Explanation:
\(E = {E_o}\cos \omega t\) \(\therefore \quad \omega = 50\pi \) \(2\pi f = 50\pi \quad \Rightarrow \quad f = 25Hz\) In one cycle ac current becomes zero twice therefore, 50 times the current becomes zero in \(1s\).
PHXII07:ALTERNATING CURRENT
356052
An \(ac\) generator produced an output voltage \(E = 170\sin (377t)\) volts, where \(t\) is in seconds. The frequency of \(ac\) voltage is
1 \(50\,Hz\)
2 \(110\,Hz\)
3 \(60\,Hz\)
4 \(230\,Hz\)
Explanation:
\(2\pi v = 377 \Rightarrow v = 60.03Hz\)
PHXII07:ALTERNATING CURRENT
356053
The peak voltage of \(440\;Volt\) \(A C\) mains is equal to (in \(Volt\))
1 155.6
2 220.0
3 622.0
4 440.0
Explanation:
\(V_{0}=\sqrt{2} \cdot V_{\text {r.m.s }}=\sqrt{2} \times 440=622\) \(Volt\) So correct option is (3).
356050
The peak value of an alternating \(e.m.f.\) \({E}\) is given by \({E=E_{0} \cos \omega t}\) is \(10\,volts\) and its frequency is \(50\,Hz\). At time \({t=\dfrac{1}{600} s}\), the instantaneous emf is
356051
An \(ac\) voltage is represented by \(E = 220\sqrt 2 \cos (50\pi )t\) How many times will the current becomes zero in 1\(s\)?
1 \({\rm{100}}\,{\rm{times}}\)
2 \({\rm{50}}\,{\rm{times}}\)
3 \({\rm{25}}\,{\rm{times}}\)
4 \({\rm{30}}\,{\rm{times}}\)
Explanation:
\(E = {E_o}\cos \omega t\) \(\therefore \quad \omega = 50\pi \) \(2\pi f = 50\pi \quad \Rightarrow \quad f = 25Hz\) In one cycle ac current becomes zero twice therefore, 50 times the current becomes zero in \(1s\).
PHXII07:ALTERNATING CURRENT
356052
An \(ac\) generator produced an output voltage \(E = 170\sin (377t)\) volts, where \(t\) is in seconds. The frequency of \(ac\) voltage is
1 \(50\,Hz\)
2 \(110\,Hz\)
3 \(60\,Hz\)
4 \(230\,Hz\)
Explanation:
\(2\pi v = 377 \Rightarrow v = 60.03Hz\)
PHXII07:ALTERNATING CURRENT
356053
The peak voltage of \(440\;Volt\) \(A C\) mains is equal to (in \(Volt\))
1 155.6
2 220.0
3 622.0
4 440.0
Explanation:
\(V_{0}=\sqrt{2} \cdot V_{\text {r.m.s }}=\sqrt{2} \times 440=622\) \(Volt\) So correct option is (3).
356050
The peak value of an alternating \(e.m.f.\) \({E}\) is given by \({E=E_{0} \cos \omega t}\) is \(10\,volts\) and its frequency is \(50\,Hz\). At time \({t=\dfrac{1}{600} s}\), the instantaneous emf is
356051
An \(ac\) voltage is represented by \(E = 220\sqrt 2 \cos (50\pi )t\) How many times will the current becomes zero in 1\(s\)?
1 \({\rm{100}}\,{\rm{times}}\)
2 \({\rm{50}}\,{\rm{times}}\)
3 \({\rm{25}}\,{\rm{times}}\)
4 \({\rm{30}}\,{\rm{times}}\)
Explanation:
\(E = {E_o}\cos \omega t\) \(\therefore \quad \omega = 50\pi \) \(2\pi f = 50\pi \quad \Rightarrow \quad f = 25Hz\) In one cycle ac current becomes zero twice therefore, 50 times the current becomes zero in \(1s\).
PHXII07:ALTERNATING CURRENT
356052
An \(ac\) generator produced an output voltage \(E = 170\sin (377t)\) volts, where \(t\) is in seconds. The frequency of \(ac\) voltage is
1 \(50\,Hz\)
2 \(110\,Hz\)
3 \(60\,Hz\)
4 \(230\,Hz\)
Explanation:
\(2\pi v = 377 \Rightarrow v = 60.03Hz\)
PHXII07:ALTERNATING CURRENT
356053
The peak voltage of \(440\;Volt\) \(A C\) mains is equal to (in \(Volt\))
1 155.6
2 220.0
3 622.0
4 440.0
Explanation:
\(V_{0}=\sqrt{2} \cdot V_{\text {r.m.s }}=\sqrt{2} \times 440=622\) \(Volt\) So correct option is (3).
356050
The peak value of an alternating \(e.m.f.\) \({E}\) is given by \({E=E_{0} \cos \omega t}\) is \(10\,volts\) and its frequency is \(50\,Hz\). At time \({t=\dfrac{1}{600} s}\), the instantaneous emf is
356051
An \(ac\) voltage is represented by \(E = 220\sqrt 2 \cos (50\pi )t\) How many times will the current becomes zero in 1\(s\)?
1 \({\rm{100}}\,{\rm{times}}\)
2 \({\rm{50}}\,{\rm{times}}\)
3 \({\rm{25}}\,{\rm{times}}\)
4 \({\rm{30}}\,{\rm{times}}\)
Explanation:
\(E = {E_o}\cos \omega t\) \(\therefore \quad \omega = 50\pi \) \(2\pi f = 50\pi \quad \Rightarrow \quad f = 25Hz\) In one cycle ac current becomes zero twice therefore, 50 times the current becomes zero in \(1s\).
PHXII07:ALTERNATING CURRENT
356052
An \(ac\) generator produced an output voltage \(E = 170\sin (377t)\) volts, where \(t\) is in seconds. The frequency of \(ac\) voltage is
1 \(50\,Hz\)
2 \(110\,Hz\)
3 \(60\,Hz\)
4 \(230\,Hz\)
Explanation:
\(2\pi v = 377 \Rightarrow v = 60.03Hz\)
PHXII07:ALTERNATING CURRENT
356053
The peak voltage of \(440\;Volt\) \(A C\) mains is equal to (in \(Volt\))
1 155.6
2 220.0
3 622.0
4 440.0
Explanation:
\(V_{0}=\sqrt{2} \cdot V_{\text {r.m.s }}=\sqrt{2} \times 440=622\) \(Volt\) So correct option is (3).
