155094
In a L-C-R circuit, the capacitance is made one-fourth, then what should be change in inductance, so that the circuit remains in resonance?
1 8 times
2 $1 / 4$ times
3 2 times
4 4 times
Explanation:
D In a LCR circuit, At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ or $\quad \omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $2 \pi \mathrm{f}=\frac{1}{\sqrt{\mathrm{LC}}}$ Resonance frequency $(f)=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ For the circuit to remain in resonance $\sqrt{\mathrm{LC}}=$ constant $\mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}$ $\mathrm{~L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \frac{\mathrm{C}_{1}}{4}$ $\mathrm{~L}_{2}=4 \mathrm{~L}_{1}$
JIPMER-2005
Alternating Current
155095
A coil has an inductance of $50 \mathrm{mH}$ and an ohmic resistance of $0.5 \Omega$. A $5 \mathrm{~V}$ e.m.f is applied across the coil. How much energy (in joules) is stored in the magnetic field after the current through the coil has built to its steady state value?
1 2.5
2 5.0
3 0.5
4 10.0
Explanation:
A $\mathrm{L}=50 \mathrm{mH}=50 \times 10^{-3} \mathrm{H}$ $\mathrm{E}=5 \mathrm{~V}$ $\mathrm{R}=0.5 \Omega$ $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{5}{0.5}=10 \mathrm{~A}$ We know that, energy stored- $\mathrm{U}=\frac{1}{2} \mathrm{LI}^{2}$ $\mathrm{U}=\frac{1}{2} \times 50 \times 10^{-3} \times 10 \times 10$ $\mathrm{U}=\frac{1}{2} \times 50 \times 10^{-1}$ $\mathrm{U}=2.5 \mathrm{~J}$ A.C. Circuit (L-C-R, LC Circuit)
AMU-2007
Alternating Current
155087
The energy stored in a $50 \mathrm{mH}$ inductor carrying a current of $4 \mathrm{~A}$ will be
155094
In a L-C-R circuit, the capacitance is made one-fourth, then what should be change in inductance, so that the circuit remains in resonance?
1 8 times
2 $1 / 4$ times
3 2 times
4 4 times
Explanation:
D In a LCR circuit, At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ or $\quad \omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $2 \pi \mathrm{f}=\frac{1}{\sqrt{\mathrm{LC}}}$ Resonance frequency $(f)=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ For the circuit to remain in resonance $\sqrt{\mathrm{LC}}=$ constant $\mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}$ $\mathrm{~L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \frac{\mathrm{C}_{1}}{4}$ $\mathrm{~L}_{2}=4 \mathrm{~L}_{1}$
JIPMER-2005
Alternating Current
155095
A coil has an inductance of $50 \mathrm{mH}$ and an ohmic resistance of $0.5 \Omega$. A $5 \mathrm{~V}$ e.m.f is applied across the coil. How much energy (in joules) is stored in the magnetic field after the current through the coil has built to its steady state value?
1 2.5
2 5.0
3 0.5
4 10.0
Explanation:
A $\mathrm{L}=50 \mathrm{mH}=50 \times 10^{-3} \mathrm{H}$ $\mathrm{E}=5 \mathrm{~V}$ $\mathrm{R}=0.5 \Omega$ $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{5}{0.5}=10 \mathrm{~A}$ We know that, energy stored- $\mathrm{U}=\frac{1}{2} \mathrm{LI}^{2}$ $\mathrm{U}=\frac{1}{2} \times 50 \times 10^{-3} \times 10 \times 10$ $\mathrm{U}=\frac{1}{2} \times 50 \times 10^{-1}$ $\mathrm{U}=2.5 \mathrm{~J}$ A.C. Circuit (L-C-R, LC Circuit)
AMU-2007
Alternating Current
155087
The energy stored in a $50 \mathrm{mH}$ inductor carrying a current of $4 \mathrm{~A}$ will be
155094
In a L-C-R circuit, the capacitance is made one-fourth, then what should be change in inductance, so that the circuit remains in resonance?
1 8 times
2 $1 / 4$ times
3 2 times
4 4 times
Explanation:
D In a LCR circuit, At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ or $\quad \omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $2 \pi \mathrm{f}=\frac{1}{\sqrt{\mathrm{LC}}}$ Resonance frequency $(f)=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ For the circuit to remain in resonance $\sqrt{\mathrm{LC}}=$ constant $\mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}$ $\mathrm{~L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \frac{\mathrm{C}_{1}}{4}$ $\mathrm{~L}_{2}=4 \mathrm{~L}_{1}$
JIPMER-2005
Alternating Current
155095
A coil has an inductance of $50 \mathrm{mH}$ and an ohmic resistance of $0.5 \Omega$. A $5 \mathrm{~V}$ e.m.f is applied across the coil. How much energy (in joules) is stored in the magnetic field after the current through the coil has built to its steady state value?
1 2.5
2 5.0
3 0.5
4 10.0
Explanation:
A $\mathrm{L}=50 \mathrm{mH}=50 \times 10^{-3} \mathrm{H}$ $\mathrm{E}=5 \mathrm{~V}$ $\mathrm{R}=0.5 \Omega$ $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{5}{0.5}=10 \mathrm{~A}$ We know that, energy stored- $\mathrm{U}=\frac{1}{2} \mathrm{LI}^{2}$ $\mathrm{U}=\frac{1}{2} \times 50 \times 10^{-3} \times 10 \times 10$ $\mathrm{U}=\frac{1}{2} \times 50 \times 10^{-1}$ $\mathrm{U}=2.5 \mathrm{~J}$ A.C. Circuit (L-C-R, LC Circuit)
AMU-2007
Alternating Current
155087
The energy stored in a $50 \mathrm{mH}$ inductor carrying a current of $4 \mathrm{~A}$ will be
155094
In a L-C-R circuit, the capacitance is made one-fourth, then what should be change in inductance, so that the circuit remains in resonance?
1 8 times
2 $1 / 4$ times
3 2 times
4 4 times
Explanation:
D In a LCR circuit, At resonance, $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$ or $\quad \omega=\frac{1}{\sqrt{\mathrm{LC}}}$ $2 \pi \mathrm{f}=\frac{1}{\sqrt{\mathrm{LC}}}$ Resonance frequency $(f)=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$ For the circuit to remain in resonance $\sqrt{\mathrm{LC}}=$ constant $\mathrm{L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \mathrm{C}_{2}$ $\mathrm{~L}_{1} \mathrm{C}_{1}=\mathrm{L}_{2} \frac{\mathrm{C}_{1}}{4}$ $\mathrm{~L}_{2}=4 \mathrm{~L}_{1}$
JIPMER-2005
Alternating Current
155095
A coil has an inductance of $50 \mathrm{mH}$ and an ohmic resistance of $0.5 \Omega$. A $5 \mathrm{~V}$ e.m.f is applied across the coil. How much energy (in joules) is stored in the magnetic field after the current through the coil has built to its steady state value?
1 2.5
2 5.0
3 0.5
4 10.0
Explanation:
A $\mathrm{L}=50 \mathrm{mH}=50 \times 10^{-3} \mathrm{H}$ $\mathrm{E}=5 \mathrm{~V}$ $\mathrm{R}=0.5 \Omega$ $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{5}{0.5}=10 \mathrm{~A}$ We know that, energy stored- $\mathrm{U}=\frac{1}{2} \mathrm{LI}^{2}$ $\mathrm{U}=\frac{1}{2} \times 50 \times 10^{-3} \times 10 \times 10$ $\mathrm{U}=\frac{1}{2} \times 50 \times 10^{-1}$ $\mathrm{U}=2.5 \mathrm{~J}$ A.C. Circuit (L-C-R, LC Circuit)
AMU-2007
Alternating Current
155087
The energy stored in a $50 \mathrm{mH}$ inductor carrying a current of $4 \mathrm{~A}$ will be
