356179
If maximum energy is stored in a capactior at \(t=0\), then find the time after which current in the circuit will be maximum.
1 \(\frac{\pi }{2}\,ms\)
2 \(\frac{\pi }{4}\,ms\)
3 \(\pi \,ms\)
4 \(2\;ms\)
Explanation:
Given, \(L = 25\,mH\), \( = 25 \times {10^{ - 3}}H\) \(C = 10\,\mu F = {10^{ - 5}}\;F\) If \(T\) be the time period in \(L-C\) oscillation, then \(T = 2\pi \sqrt {LC} \) \( = 2\pi \sqrt {25 \times {{10}^{ - 3}} \times {{10}^{ - 5}}} \) \( = \pi \times {10^{ - 3}}\;s = \pi \,ms\) Current in the circuit will be maximum when \(t=\dfrac{T}{4}=\dfrac{\pi}{4} m s\)
AIIMS - 2019
PHXII07:ALTERNATING CURRENT
356180
A resistance \({R}\) draws power \({P}\) when connected to an ac source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes \({Z}\), the power drawn will be
356181
A series \(LR\) circuit connected with an \(ac\) source \(E=(25 \sin 1000 t) V\) has a power factor of \(\dfrac{1}{\sqrt{2}}\). If the source of emf is changed to \(E=(20 \sin 2000 t) V\), the new power factor of the circuit will be
1 \(\dfrac{1}{\sqrt{7}}\)
2 \(\dfrac{1}{\sqrt{5}}\)
3 \(\dfrac{1}{\sqrt{2}}\)
4 \(\dfrac{1}{\sqrt{3}}\)
Explanation:
Power factor of a series \(L R\) circuit is \(\cos \phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}\) \(E_{1}=25 \sin (1000 t) V\) Given: \(\cos \phi_{1}=\dfrac{R}{\sqrt{R^{2}+(1000)^{2} L^{2}}}=\dfrac{1}{\sqrt{2}}\) \(\left( {\because {\omega _1} = 1000\,rad/s} \right)\) \(\Rightarrow \dfrac{R^{2}}{R^{2}+(1000)^{2} L^{2}}=\dfrac{1}{2}\) \(\Rightarrow R^{2}=(1000)^{2} L^{2} \Rightarrow R=1000 L\) \({E_2} = 20\sin \,(2000\,t)V\left( {\because {\omega _2} = 2000\,rad/s} \right)\) Now, \(\cos \phi_{2}=\dfrac{R}{\sqrt{R^{2}+(2000)^{2} L^{2}}}\) Substituting the value of \(R\), we have \(\cos \phi_{2}=\dfrac{1000 L}{\sqrt{(1000)^{2} L^{2}+(2000)^{2} L^{2}}}=\dfrac{1}{\sqrt{5}}\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356182
A coil of resistance \(10\, \Omega\) and an inductance \(5\,H\) is connected to a \(100\;V\) battery. The energy stored in the coil is
1 325 erg
2 \(125\;J\)
3 250 erg
4 \(250\;J\)
Explanation:
Here,\(R = 10\,\Omega ,L = 5H,V = 100\;V\) The current in the coil is \(I = \frac{V}{R} = \frac{{100\;V}}{{10\,\Omega }} = 10\;A\) The energy stored in the coil is \(U = \frac{1}{2}L{I^2} = \frac{1}{2}(5H){(10\;A)^2} = 250\;J\)
356179
If maximum energy is stored in a capactior at \(t=0\), then find the time after which current in the circuit will be maximum.
1 \(\frac{\pi }{2}\,ms\)
2 \(\frac{\pi }{4}\,ms\)
3 \(\pi \,ms\)
4 \(2\;ms\)
Explanation:
Given, \(L = 25\,mH\), \( = 25 \times {10^{ - 3}}H\) \(C = 10\,\mu F = {10^{ - 5}}\;F\) If \(T\) be the time period in \(L-C\) oscillation, then \(T = 2\pi \sqrt {LC} \) \( = 2\pi \sqrt {25 \times {{10}^{ - 3}} \times {{10}^{ - 5}}} \) \( = \pi \times {10^{ - 3}}\;s = \pi \,ms\) Current in the circuit will be maximum when \(t=\dfrac{T}{4}=\dfrac{\pi}{4} m s\)
AIIMS - 2019
PHXII07:ALTERNATING CURRENT
356180
A resistance \({R}\) draws power \({P}\) when connected to an ac source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes \({Z}\), the power drawn will be
356181
A series \(LR\) circuit connected with an \(ac\) source \(E=(25 \sin 1000 t) V\) has a power factor of \(\dfrac{1}{\sqrt{2}}\). If the source of emf is changed to \(E=(20 \sin 2000 t) V\), the new power factor of the circuit will be
1 \(\dfrac{1}{\sqrt{7}}\)
2 \(\dfrac{1}{\sqrt{5}}\)
3 \(\dfrac{1}{\sqrt{2}}\)
4 \(\dfrac{1}{\sqrt{3}}\)
Explanation:
Power factor of a series \(L R\) circuit is \(\cos \phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}\) \(E_{1}=25 \sin (1000 t) V\) Given: \(\cos \phi_{1}=\dfrac{R}{\sqrt{R^{2}+(1000)^{2} L^{2}}}=\dfrac{1}{\sqrt{2}}\) \(\left( {\because {\omega _1} = 1000\,rad/s} \right)\) \(\Rightarrow \dfrac{R^{2}}{R^{2}+(1000)^{2} L^{2}}=\dfrac{1}{2}\) \(\Rightarrow R^{2}=(1000)^{2} L^{2} \Rightarrow R=1000 L\) \({E_2} = 20\sin \,(2000\,t)V\left( {\because {\omega _2} = 2000\,rad/s} \right)\) Now, \(\cos \phi_{2}=\dfrac{R}{\sqrt{R^{2}+(2000)^{2} L^{2}}}\) Substituting the value of \(R\), we have \(\cos \phi_{2}=\dfrac{1000 L}{\sqrt{(1000)^{2} L^{2}+(2000)^{2} L^{2}}}=\dfrac{1}{\sqrt{5}}\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356182
A coil of resistance \(10\, \Omega\) and an inductance \(5\,H\) is connected to a \(100\;V\) battery. The energy stored in the coil is
1 325 erg
2 \(125\;J\)
3 250 erg
4 \(250\;J\)
Explanation:
Here,\(R = 10\,\Omega ,L = 5H,V = 100\;V\) The current in the coil is \(I = \frac{V}{R} = \frac{{100\;V}}{{10\,\Omega }} = 10\;A\) The energy stored in the coil is \(U = \frac{1}{2}L{I^2} = \frac{1}{2}(5H){(10\;A)^2} = 250\;J\)
356179
If maximum energy is stored in a capactior at \(t=0\), then find the time after which current in the circuit will be maximum.
1 \(\frac{\pi }{2}\,ms\)
2 \(\frac{\pi }{4}\,ms\)
3 \(\pi \,ms\)
4 \(2\;ms\)
Explanation:
Given, \(L = 25\,mH\), \( = 25 \times {10^{ - 3}}H\) \(C = 10\,\mu F = {10^{ - 5}}\;F\) If \(T\) be the time period in \(L-C\) oscillation, then \(T = 2\pi \sqrt {LC} \) \( = 2\pi \sqrt {25 \times {{10}^{ - 3}} \times {{10}^{ - 5}}} \) \( = \pi \times {10^{ - 3}}\;s = \pi \,ms\) Current in the circuit will be maximum when \(t=\dfrac{T}{4}=\dfrac{\pi}{4} m s\)
AIIMS - 2019
PHXII07:ALTERNATING CURRENT
356180
A resistance \({R}\) draws power \({P}\) when connected to an ac source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes \({Z}\), the power drawn will be
356181
A series \(LR\) circuit connected with an \(ac\) source \(E=(25 \sin 1000 t) V\) has a power factor of \(\dfrac{1}{\sqrt{2}}\). If the source of emf is changed to \(E=(20 \sin 2000 t) V\), the new power factor of the circuit will be
1 \(\dfrac{1}{\sqrt{7}}\)
2 \(\dfrac{1}{\sqrt{5}}\)
3 \(\dfrac{1}{\sqrt{2}}\)
4 \(\dfrac{1}{\sqrt{3}}\)
Explanation:
Power factor of a series \(L R\) circuit is \(\cos \phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}\) \(E_{1}=25 \sin (1000 t) V\) Given: \(\cos \phi_{1}=\dfrac{R}{\sqrt{R^{2}+(1000)^{2} L^{2}}}=\dfrac{1}{\sqrt{2}}\) \(\left( {\because {\omega _1} = 1000\,rad/s} \right)\) \(\Rightarrow \dfrac{R^{2}}{R^{2}+(1000)^{2} L^{2}}=\dfrac{1}{2}\) \(\Rightarrow R^{2}=(1000)^{2} L^{2} \Rightarrow R=1000 L\) \({E_2} = 20\sin \,(2000\,t)V\left( {\because {\omega _2} = 2000\,rad/s} \right)\) Now, \(\cos \phi_{2}=\dfrac{R}{\sqrt{R^{2}+(2000)^{2} L^{2}}}\) Substituting the value of \(R\), we have \(\cos \phi_{2}=\dfrac{1000 L}{\sqrt{(1000)^{2} L^{2}+(2000)^{2} L^{2}}}=\dfrac{1}{\sqrt{5}}\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356182
A coil of resistance \(10\, \Omega\) and an inductance \(5\,H\) is connected to a \(100\;V\) battery. The energy stored in the coil is
1 325 erg
2 \(125\;J\)
3 250 erg
4 \(250\;J\)
Explanation:
Here,\(R = 10\,\Omega ,L = 5H,V = 100\;V\) The current in the coil is \(I = \frac{V}{R} = \frac{{100\;V}}{{10\,\Omega }} = 10\;A\) The energy stored in the coil is \(U = \frac{1}{2}L{I^2} = \frac{1}{2}(5H){(10\;A)^2} = 250\;J\)
356179
If maximum energy is stored in a capactior at \(t=0\), then find the time after which current in the circuit will be maximum.
1 \(\frac{\pi }{2}\,ms\)
2 \(\frac{\pi }{4}\,ms\)
3 \(\pi \,ms\)
4 \(2\;ms\)
Explanation:
Given, \(L = 25\,mH\), \( = 25 \times {10^{ - 3}}H\) \(C = 10\,\mu F = {10^{ - 5}}\;F\) If \(T\) be the time period in \(L-C\) oscillation, then \(T = 2\pi \sqrt {LC} \) \( = 2\pi \sqrt {25 \times {{10}^{ - 3}} \times {{10}^{ - 5}}} \) \( = \pi \times {10^{ - 3}}\;s = \pi \,ms\) Current in the circuit will be maximum when \(t=\dfrac{T}{4}=\dfrac{\pi}{4} m s\)
AIIMS - 2019
PHXII07:ALTERNATING CURRENT
356180
A resistance \({R}\) draws power \({P}\) when connected to an ac source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes \({Z}\), the power drawn will be
356181
A series \(LR\) circuit connected with an \(ac\) source \(E=(25 \sin 1000 t) V\) has a power factor of \(\dfrac{1}{\sqrt{2}}\). If the source of emf is changed to \(E=(20 \sin 2000 t) V\), the new power factor of the circuit will be
1 \(\dfrac{1}{\sqrt{7}}\)
2 \(\dfrac{1}{\sqrt{5}}\)
3 \(\dfrac{1}{\sqrt{2}}\)
4 \(\dfrac{1}{\sqrt{3}}\)
Explanation:
Power factor of a series \(L R\) circuit is \(\cos \phi=\dfrac{R}{Z}=\dfrac{R}{\sqrt{R^{2}+\omega^{2} L^{2}}}\) \(E_{1}=25 \sin (1000 t) V\) Given: \(\cos \phi_{1}=\dfrac{R}{\sqrt{R^{2}+(1000)^{2} L^{2}}}=\dfrac{1}{\sqrt{2}}\) \(\left( {\because {\omega _1} = 1000\,rad/s} \right)\) \(\Rightarrow \dfrac{R^{2}}{R^{2}+(1000)^{2} L^{2}}=\dfrac{1}{2}\) \(\Rightarrow R^{2}=(1000)^{2} L^{2} \Rightarrow R=1000 L\) \({E_2} = 20\sin \,(2000\,t)V\left( {\because {\omega _2} = 2000\,rad/s} \right)\) Now, \(\cos \phi_{2}=\dfrac{R}{\sqrt{R^{2}+(2000)^{2} L^{2}}}\) Substituting the value of \(R\), we have \(\cos \phi_{2}=\dfrac{1000 L}{\sqrt{(1000)^{2} L^{2}+(2000)^{2} L^{2}}}=\dfrac{1}{\sqrt{5}}\)
JEE - 2024
PHXII07:ALTERNATING CURRENT
356182
A coil of resistance \(10\, \Omega\) and an inductance \(5\,H\) is connected to a \(100\;V\) battery. The energy stored in the coil is
1 325 erg
2 \(125\;J\)
3 250 erg
4 \(250\;J\)
Explanation:
Here,\(R = 10\,\Omega ,L = 5H,V = 100\;V\) The current in the coil is \(I = \frac{V}{R} = \frac{{100\;V}}{{10\,\Omega }} = 10\;A\) The energy stored in the coil is \(U = \frac{1}{2}L{I^2} = \frac{1}{2}(5H){(10\;A)^2} = 250\;J\)
