356046
The peak value of \(220\,volts,\) of \(a.c\) mains is:
1 \(155.6\;V\)
2 \(200\;V\)
3 \(311\;V\)
4 \(440\;V\)
Explanation:
\(V_{r m s}=\dfrac{V_{0}}{\sqrt{2}}\) \(\Rightarrow\) Peak value is \(V_{0}=\sqrt{2} \times V_{r m s}\) \(=1.41 \times 220\) \({V_0} = 311\;V\) So correct option is (3).
PHXII07:ALTERNATING CURRENT
356047
The instantaneous value of current in an \(A\).\(C\). circuit is \(I = 2\sin (100\pi t + \pi /3)A.\)The current will be maximum for the first time at
1 \(t = \frac{1}{{600}}s\)
2 \(t = \frac{1}{{200}}s\)
3 \(t = \frac{1}{{100}}s\)
4 \(t = \frac{1}{{400}}s\)
Explanation:
Current will be max at first time when \(100\pi t + \pi /3 = \pi /2\)\( \Rightarrow 100\pi t = \pi /6\)\( \Rightarrow t = 1/600s\)
PHXII07:ALTERNATING CURRENT
356048
The \(rms\) value of an \(ac\) of 50 \(Hz\) is 10\(A\). The time taken by an alternating current in reaching from zero to maximum value or the peak value will be
1 \(5 \times {10^{ - 3}}\,s\,{\rm{and}}\,14.14A\)
2 \(1 \times {10^{ - 2}}s\,\,{\rm{and}}\,7.07A\)
3 \(2 \times {10^{ - 2}}s\,{\rm{and}}\,14.14A\)
4 \(5 \times {10^{ - 3}}s\,{\rm{and}}\,7.07A\)
Explanation:
Time for reaching maximum or peak value from 0 \( = \frac{T}{4} = \frac{1}{4} \times \frac{1}{{50}}s = \frac{1}{{200}}s = 5 \times {10^{ - 3}}s\) \({I_o} = 10\sqrt 2 A = 14.14A\)
PHXII07:ALTERNATING CURRENT
356049
The peak voltage of the \(ac\) source is equal to
1 The rms value of the ac source
2 \(\sqrt 2 \) times the rms value of the ac source
3 \(1/\sqrt 2 \) time the rms value fo the ac source
4 the value of voltage supplied to the circuit.
Explanation:
Peak voltage is \(\sqrt 2 \) times rms voltage in alternating current.
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PHXII07:ALTERNATING CURRENT
356046
The peak value of \(220\,volts,\) of \(a.c\) mains is:
1 \(155.6\;V\)
2 \(200\;V\)
3 \(311\;V\)
4 \(440\;V\)
Explanation:
\(V_{r m s}=\dfrac{V_{0}}{\sqrt{2}}\) \(\Rightarrow\) Peak value is \(V_{0}=\sqrt{2} \times V_{r m s}\) \(=1.41 \times 220\) \({V_0} = 311\;V\) So correct option is (3).
PHXII07:ALTERNATING CURRENT
356047
The instantaneous value of current in an \(A\).\(C\). circuit is \(I = 2\sin (100\pi t + \pi /3)A.\)The current will be maximum for the first time at
1 \(t = \frac{1}{{600}}s\)
2 \(t = \frac{1}{{200}}s\)
3 \(t = \frac{1}{{100}}s\)
4 \(t = \frac{1}{{400}}s\)
Explanation:
Current will be max at first time when \(100\pi t + \pi /3 = \pi /2\)\( \Rightarrow 100\pi t = \pi /6\)\( \Rightarrow t = 1/600s\)
PHXII07:ALTERNATING CURRENT
356048
The \(rms\) value of an \(ac\) of 50 \(Hz\) is 10\(A\). The time taken by an alternating current in reaching from zero to maximum value or the peak value will be
1 \(5 \times {10^{ - 3}}\,s\,{\rm{and}}\,14.14A\)
2 \(1 \times {10^{ - 2}}s\,\,{\rm{and}}\,7.07A\)
3 \(2 \times {10^{ - 2}}s\,{\rm{and}}\,14.14A\)
4 \(5 \times {10^{ - 3}}s\,{\rm{and}}\,7.07A\)
Explanation:
Time for reaching maximum or peak value from 0 \( = \frac{T}{4} = \frac{1}{4} \times \frac{1}{{50}}s = \frac{1}{{200}}s = 5 \times {10^{ - 3}}s\) \({I_o} = 10\sqrt 2 A = 14.14A\)
PHXII07:ALTERNATING CURRENT
356049
The peak voltage of the \(ac\) source is equal to
1 The rms value of the ac source
2 \(\sqrt 2 \) times the rms value of the ac source
3 \(1/\sqrt 2 \) time the rms value fo the ac source
4 the value of voltage supplied to the circuit.
Explanation:
Peak voltage is \(\sqrt 2 \) times rms voltage in alternating current.
356046
The peak value of \(220\,volts,\) of \(a.c\) mains is:
1 \(155.6\;V\)
2 \(200\;V\)
3 \(311\;V\)
4 \(440\;V\)
Explanation:
\(V_{r m s}=\dfrac{V_{0}}{\sqrt{2}}\) \(\Rightarrow\) Peak value is \(V_{0}=\sqrt{2} \times V_{r m s}\) \(=1.41 \times 220\) \({V_0} = 311\;V\) So correct option is (3).
PHXII07:ALTERNATING CURRENT
356047
The instantaneous value of current in an \(A\).\(C\). circuit is \(I = 2\sin (100\pi t + \pi /3)A.\)The current will be maximum for the first time at
1 \(t = \frac{1}{{600}}s\)
2 \(t = \frac{1}{{200}}s\)
3 \(t = \frac{1}{{100}}s\)
4 \(t = \frac{1}{{400}}s\)
Explanation:
Current will be max at first time when \(100\pi t + \pi /3 = \pi /2\)\( \Rightarrow 100\pi t = \pi /6\)\( \Rightarrow t = 1/600s\)
PHXII07:ALTERNATING CURRENT
356048
The \(rms\) value of an \(ac\) of 50 \(Hz\) is 10\(A\). The time taken by an alternating current in reaching from zero to maximum value or the peak value will be
1 \(5 \times {10^{ - 3}}\,s\,{\rm{and}}\,14.14A\)
2 \(1 \times {10^{ - 2}}s\,\,{\rm{and}}\,7.07A\)
3 \(2 \times {10^{ - 2}}s\,{\rm{and}}\,14.14A\)
4 \(5 \times {10^{ - 3}}s\,{\rm{and}}\,7.07A\)
Explanation:
Time for reaching maximum or peak value from 0 \( = \frac{T}{4} = \frac{1}{4} \times \frac{1}{{50}}s = \frac{1}{{200}}s = 5 \times {10^{ - 3}}s\) \({I_o} = 10\sqrt 2 A = 14.14A\)
PHXII07:ALTERNATING CURRENT
356049
The peak voltage of the \(ac\) source is equal to
1 The rms value of the ac source
2 \(\sqrt 2 \) times the rms value of the ac source
3 \(1/\sqrt 2 \) time the rms value fo the ac source
4 the value of voltage supplied to the circuit.
Explanation:
Peak voltage is \(\sqrt 2 \) times rms voltage in alternating current.
356046
The peak value of \(220\,volts,\) of \(a.c\) mains is:
1 \(155.6\;V\)
2 \(200\;V\)
3 \(311\;V\)
4 \(440\;V\)
Explanation:
\(V_{r m s}=\dfrac{V_{0}}{\sqrt{2}}\) \(\Rightarrow\) Peak value is \(V_{0}=\sqrt{2} \times V_{r m s}\) \(=1.41 \times 220\) \({V_0} = 311\;V\) So correct option is (3).
PHXII07:ALTERNATING CURRENT
356047
The instantaneous value of current in an \(A\).\(C\). circuit is \(I = 2\sin (100\pi t + \pi /3)A.\)The current will be maximum for the first time at
1 \(t = \frac{1}{{600}}s\)
2 \(t = \frac{1}{{200}}s\)
3 \(t = \frac{1}{{100}}s\)
4 \(t = \frac{1}{{400}}s\)
Explanation:
Current will be max at first time when \(100\pi t + \pi /3 = \pi /2\)\( \Rightarrow 100\pi t = \pi /6\)\( \Rightarrow t = 1/600s\)
PHXII07:ALTERNATING CURRENT
356048
The \(rms\) value of an \(ac\) of 50 \(Hz\) is 10\(A\). The time taken by an alternating current in reaching from zero to maximum value or the peak value will be
1 \(5 \times {10^{ - 3}}\,s\,{\rm{and}}\,14.14A\)
2 \(1 \times {10^{ - 2}}s\,\,{\rm{and}}\,7.07A\)
3 \(2 \times {10^{ - 2}}s\,{\rm{and}}\,14.14A\)
4 \(5 \times {10^{ - 3}}s\,{\rm{and}}\,7.07A\)
Explanation:
Time for reaching maximum or peak value from 0 \( = \frac{T}{4} = \frac{1}{4} \times \frac{1}{{50}}s = \frac{1}{{200}}s = 5 \times {10^{ - 3}}s\) \({I_o} = 10\sqrt 2 A = 14.14A\)
PHXII07:ALTERNATING CURRENT
356049
The peak voltage of the \(ac\) source is equal to
1 The rms value of the ac source
2 \(\sqrt 2 \) times the rms value of the ac source
3 \(1/\sqrt 2 \) time the rms value fo the ac source
4 the value of voltage supplied to the circuit.
Explanation:
Peak voltage is \(\sqrt 2 \) times rms voltage in alternating current.