356192
In an ac circuit, \(V\) and \(I\) are given by \(V = 100\sin (100t)volts,\) \(I = 100\sin \left( {100t + \frac{\pi }{3}} \right)mA\). The power dissipated in the circuit is
356193
In an \(AC\) circuit the emf \((V)\) and the current \((i)\) at any instant are given respectively by \(V = {V_0}\,\sin \,\omega t\),\(i = {i_0}\sin \left( {\omega t - \phi } \right)\). The average power in the circuit over one cycle of \(AC\) is
356194
A \(100\,\Omega \) electric iron is connected to \(200\;V\), \(50\;Hz\,ac\) source. The average power delivered to iron, peak power and energy spent in one minute will be
1 \(400\;W,800\;W,12 \times {10^5}\;J\)
2 \(400\;W,900\;W,1.2 \times {10^5}\;J\)
3 \(500\;W,800\;W,6 \times {10^5}\;J\)
4 \(400\;W,900\;W,60 \times {10^5}\;J\)
Explanation:
Here, \(R = 100\Omega ,{V_{rms}} = 200\;V,f = 50\;Hz\) \(\langle P\rangle=\) Average power \(/\) cycle \(=V_{r m s} I_{r m s}=\dfrac{V_{r m s}^{2}}{R}\) \(\langle P\rangle=\dfrac{(200)^{2}}{100}=400\) \(watt\) \(P_{0}=\) Peak power \(/\) cycle \(=V_{0} I_{0}=\sqrt{2} V_{r m s} \times \sqrt{2} I_{r m s}\) \( \Rightarrow {P_0} = 2 \times 400 = 800\;W\) But, the frequency is \(50\;Hz\) \(\Rightarrow\) Average power spent in one second \(=\langle P\rangle f=400 \times 50\) \(\therefore\) Energy spent in one minute \(=U=\langle P\rangle f(60)\) \( \Rightarrow U = 400 \times 50 \times 60 = 12 \times {10^5}\;J\). So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356195
In an \(AC\) circuit, the voltage applied is \(E = {E_0}\sin \omega t.\) The resulting current in the circuit is \(I = {I_0}\sin \left( {\omega t - \pi /2} \right).\) The power consumption in the circuit is given by
356192
In an ac circuit, \(V\) and \(I\) are given by \(V = 100\sin (100t)volts,\) \(I = 100\sin \left( {100t + \frac{\pi }{3}} \right)mA\). The power dissipated in the circuit is
356193
In an \(AC\) circuit the emf \((V)\) and the current \((i)\) at any instant are given respectively by \(V = {V_0}\,\sin \,\omega t\),\(i = {i_0}\sin \left( {\omega t - \phi } \right)\). The average power in the circuit over one cycle of \(AC\) is
356194
A \(100\,\Omega \) electric iron is connected to \(200\;V\), \(50\;Hz\,ac\) source. The average power delivered to iron, peak power and energy spent in one minute will be
1 \(400\;W,800\;W,12 \times {10^5}\;J\)
2 \(400\;W,900\;W,1.2 \times {10^5}\;J\)
3 \(500\;W,800\;W,6 \times {10^5}\;J\)
4 \(400\;W,900\;W,60 \times {10^5}\;J\)
Explanation:
Here, \(R = 100\Omega ,{V_{rms}} = 200\;V,f = 50\;Hz\) \(\langle P\rangle=\) Average power \(/\) cycle \(=V_{r m s} I_{r m s}=\dfrac{V_{r m s}^{2}}{R}\) \(\langle P\rangle=\dfrac{(200)^{2}}{100}=400\) \(watt\) \(P_{0}=\) Peak power \(/\) cycle \(=V_{0} I_{0}=\sqrt{2} V_{r m s} \times \sqrt{2} I_{r m s}\) \( \Rightarrow {P_0} = 2 \times 400 = 800\;W\) But, the frequency is \(50\;Hz\) \(\Rightarrow\) Average power spent in one second \(=\langle P\rangle f=400 \times 50\) \(\therefore\) Energy spent in one minute \(=U=\langle P\rangle f(60)\) \( \Rightarrow U = 400 \times 50 \times 60 = 12 \times {10^5}\;J\). So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356195
In an \(AC\) circuit, the voltage applied is \(E = {E_0}\sin \omega t.\) The resulting current in the circuit is \(I = {I_0}\sin \left( {\omega t - \pi /2} \right).\) The power consumption in the circuit is given by
356192
In an ac circuit, \(V\) and \(I\) are given by \(V = 100\sin (100t)volts,\) \(I = 100\sin \left( {100t + \frac{\pi }{3}} \right)mA\). The power dissipated in the circuit is
356193
In an \(AC\) circuit the emf \((V)\) and the current \((i)\) at any instant are given respectively by \(V = {V_0}\,\sin \,\omega t\),\(i = {i_0}\sin \left( {\omega t - \phi } \right)\). The average power in the circuit over one cycle of \(AC\) is
356194
A \(100\,\Omega \) electric iron is connected to \(200\;V\), \(50\;Hz\,ac\) source. The average power delivered to iron, peak power and energy spent in one minute will be
1 \(400\;W,800\;W,12 \times {10^5}\;J\)
2 \(400\;W,900\;W,1.2 \times {10^5}\;J\)
3 \(500\;W,800\;W,6 \times {10^5}\;J\)
4 \(400\;W,900\;W,60 \times {10^5}\;J\)
Explanation:
Here, \(R = 100\Omega ,{V_{rms}} = 200\;V,f = 50\;Hz\) \(\langle P\rangle=\) Average power \(/\) cycle \(=V_{r m s} I_{r m s}=\dfrac{V_{r m s}^{2}}{R}\) \(\langle P\rangle=\dfrac{(200)^{2}}{100}=400\) \(watt\) \(P_{0}=\) Peak power \(/\) cycle \(=V_{0} I_{0}=\sqrt{2} V_{r m s} \times \sqrt{2} I_{r m s}\) \( \Rightarrow {P_0} = 2 \times 400 = 800\;W\) But, the frequency is \(50\;Hz\) \(\Rightarrow\) Average power spent in one second \(=\langle P\rangle f=400 \times 50\) \(\therefore\) Energy spent in one minute \(=U=\langle P\rangle f(60)\) \( \Rightarrow U = 400 \times 50 \times 60 = 12 \times {10^5}\;J\). So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356195
In an \(AC\) circuit, the voltage applied is \(E = {E_0}\sin \omega t.\) The resulting current in the circuit is \(I = {I_0}\sin \left( {\omega t - \pi /2} \right).\) The power consumption in the circuit is given by
356192
In an ac circuit, \(V\) and \(I\) are given by \(V = 100\sin (100t)volts,\) \(I = 100\sin \left( {100t + \frac{\pi }{3}} \right)mA\). The power dissipated in the circuit is
356193
In an \(AC\) circuit the emf \((V)\) and the current \((i)\) at any instant are given respectively by \(V = {V_0}\,\sin \,\omega t\),\(i = {i_0}\sin \left( {\omega t - \phi } \right)\). The average power in the circuit over one cycle of \(AC\) is
356194
A \(100\,\Omega \) electric iron is connected to \(200\;V\), \(50\;Hz\,ac\) source. The average power delivered to iron, peak power and energy spent in one minute will be
1 \(400\;W,800\;W,12 \times {10^5}\;J\)
2 \(400\;W,900\;W,1.2 \times {10^5}\;J\)
3 \(500\;W,800\;W,6 \times {10^5}\;J\)
4 \(400\;W,900\;W,60 \times {10^5}\;J\)
Explanation:
Here, \(R = 100\Omega ,{V_{rms}} = 200\;V,f = 50\;Hz\) \(\langle P\rangle=\) Average power \(/\) cycle \(=V_{r m s} I_{r m s}=\dfrac{V_{r m s}^{2}}{R}\) \(\langle P\rangle=\dfrac{(200)^{2}}{100}=400\) \(watt\) \(P_{0}=\) Peak power \(/\) cycle \(=V_{0} I_{0}=\sqrt{2} V_{r m s} \times \sqrt{2} I_{r m s}\) \( \Rightarrow {P_0} = 2 \times 400 = 800\;W\) But, the frequency is \(50\;Hz\) \(\Rightarrow\) Average power spent in one second \(=\langle P\rangle f=400 \times 50\) \(\therefore\) Energy spent in one minute \(=U=\langle P\rangle f(60)\) \( \Rightarrow U = 400 \times 50 \times 60 = 12 \times {10^5}\;J\). So, correct option is (1).
PHXII07:ALTERNATING CURRENT
356195
In an \(AC\) circuit, the voltage applied is \(E = {E_0}\sin \omega t.\) The resulting current in the circuit is \(I = {I_0}\sin \left( {\omega t - \pi /2} \right).\) The power consumption in the circuit is given by
