Combination of two identical capacitors, a resistor R and a dc voltage source of voltage is used in

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• Correct Answer: C

  Solution :

  [c] Charge on he capacitor at any time t is given by \[q=CV\text{ (1}-{{e}^{-t/\tau }})\,at\,t=2\tau \] \[q=CV\text{ (1-}{{\text{e}}^{-2}}\text{)}\]

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• Correct Answer: A

  Solution :

  [a] The equivalent primary load is \[{{R}_{1}}={{\left( \frac{{{N}_{1}}}{{{N}_{2}}} \right)}^{2}}\,{{R}_{2}}={{\left( \frac{20}{1} \right)}^{2}}\,(6.0)\,=2400\,\Omega \] Current in the primary coil  \[=\frac{240}{{{R}_{1}}}=\frac{240}{2400}=0.1A\]

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• Correct Answer: C

  Solution :

  [c] \[\frac{{{I}_{s}}}{{{I}_{p}}}=\frac{{{n}_{p}}}{{{n}_{s}}};\,\frac{80}{{{I}_{p}}}=\frac{20}{1}\] or \[{{I}_{p}}=4\,amp.\]

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• Correct Answer: A

  Solution :

  [a] Since \[\frac{{{V}_{s}}}{{{V}_{p}}}=\frac{{{N}_{s}}}{{{N}_{p}}}\] Where \[{{N}_{s}}\] = No. of turns across primary coil = 50 \[{{N}_{p}}\] = No. of turns across secondary coil = 1500 and \[{{V}_{p}}=\frac{d\phi }{dt}=\frac{d}{dt}\,({{\phi }_{0}}+4t)=\,4\Rightarrow \,V=\frac{1500}{50}\times 4\] =120 V   

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• Correct Answer: B

  Solution :

  [b] \[P={{V}_{1}}{{i}_{1}}={{V}_{2}}{{i}_{2}}\] \[{{V}_{2}}=\frac{{{10}^{4}}}{25}\] Now \[{{V}_{1}}=\frac{{{n}_{1}}}{{{n}_{2}}}\times {{V}_{2}}=\frac{8}{1}\times \frac{{{10}^{4}}}{25}\,V\]

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• Correct Answer: B

  Solution :

  [b] Power of source \[=El=240\times 0.7=166\] Efficiency \[\Rightarrow \,\,Efficiency\,=\frac{140}{166}\Rightarrow \,\eta \,=83.3%\]

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• Correct Answer: A

  Solution :

  [a] Time period, \[T=2\pi \sqrt{LC}=2\pi \sqrt{(50\times {{10}^{-3}})\times 4\times {{10}^{-\,6}}}\] \[=28\times {{10}^{-\,4}}s\] Time taken by capacitor to charge fulley, \[t=\frac{T}{4}=7\times {{10}^{-\,4}}s.\]

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• Correct Answer: A

  Solution :

  [a] \[{{N}_{P}}=400,\,n={{N}_{S}}=2000\,{{V}_{S}}=1000\,V.\] \[\frac{{{V}_{p}}}{{{V}_{S}}}=\frac{{{N}_{p}}}{{{N}_{s}}}\]of \[{{V}_{p}}=\frac{{{V}_{S}}\times {{N}_{P}}}{{{N}_{S}}}=\frac{1000\times 400}{2000}=200\,V\]

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• Correct Answer: C

  Solution :

  [c] Time period, \[{{T}_{1}}=2\pi \sqrt{LC},\,\,\,\,{{T}_{2}}=2\pi \sqrt{\frac{LC}{2}}\] \[{{T}_{3}}=2\pi \sqrt{2LC}\] Clearly \[{{t}_{2}}<{{t}_{1}}<{{t}_{3}}.\]

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• Correct Answer: D

  Solution :

  [d] A transformer is essentially an A C device. DC source so no mutual induction between coils \[\Rightarrow \,\,{{E}_{2}}=0\]and \[{{I}_{2}}=0\]

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• Correct Answer: B

  Solution :

  [b] Efficiency \[\eta =\frac{{{V}_{s}}{{I}_{s}}}{{{V}_{p}}{{I}_{p}}}\Rightarrow \,0.9\,=\frac{{{V}_{s}}(6)}{3\times {{10}^{3}}}\Rightarrow \,{{V}_{s}}\] = 450V As \[{{V}_{P}}{{I}_{p}}=3000\] So \[{{I}_{p}}=\frac{3000}{{{V}_{p}}}=\frac{3000}{200}=A=15A\]

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• Correct Answer: C

  Solution :

  [c] The transmission is undamped, because energy loss becomes negligible.

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• Correct Answer: B

  Solution :

  [b] \[{{U}_{e}}+{{U}_{m}}=160\,\,\,\,\,\,\,\,\therefore \,{{U}_{m}}=160\,-{{U}_{e}}=160-10\] \[=150\,\mu J.\]

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• Correct Answer: B

  Solution :

  [b] \[\frac{{{n}_{p}}}{{{n}_{s}}}=\frac{{{E}_{p}}}{{{E}_{s}}}=\frac{1}{25}\]                      \[\therefore \,\,{{E}_{s}}=25{{E}_{p}}\] But \[{{E}_{s}}{{I}_{s}}={{E}_{p}}{{I}_{p}}\Rightarrow \frac{{{E}_{S}}\times {{I}_{S}}}{{{E}_{p}}}\Rightarrow {{I}_{p}}=50\text{ }A\]

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• Correct Answer: A

  Solution :

  [a] \[L=10\,\,mHz={{10}^{-2}}\,Hz\] \[f=1\,MHz={{10}^{6}}\,Hz\] \[f=\frac{1}{2\pi \,\sqrt{LC}}\Rightarrow \,\,\,\,\,\,\,{{f}^{2}}=\frac{1}{4{{\pi }^{2}}LC}\]        \[\Rightarrow \,\,C=\frac{1}{4{{\pi }^{2}}{{f}^{2}}L}=\frac{1}{4\times 10\times {{10}^{-2}}\times {{10}^{12}}}\]

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• Correct Answer: C

  Solution :

  [c] For a transformer, \[{{V}_{p}}{{I}_{p}}={{V}_{s}}{{I}_{s}}\,\,\,\Rightarrow \,\,{{I}_{S}}=\frac{{{V}_{p}}{{I}_{p}}}{{{V}_{s}}}=\frac{4000\times 100}{240000}A\] =1.67A

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• Correct Answer: B

  Solution :

  [b] We know that frequency of electrical oscillation in L.C. circuit is \[f=\frac{1}{2\pi }\sqrt{\frac{1}{LC}}\] Now, L=2L & C=4C \[f'=\frac{1}{2\pi }\sqrt{\frac{1}{2L\cdot 4C}}=\frac{1}{2\pi }\sqrt{\frac{1}{LC}}\times \frac{1}{2\sqrt{2}}\Rightarrow \,f'\] \[=\frac{1}{2\sqrt{2}}\times f\]

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• Correct Answer: D

  Solution :

  [d] \[\frac{{{V}_{2}}}{{{V}_{1}}}=0.8\frac{{{I}_{1}}}{{{I}_{2}}}\,\Rightarrow \,\frac{{{V}_{2}}{{I}_{2}}}{{{V}_{1}}{{I}_{1}}}=0.8\] \[{{V}_{1}}=220\,V,\,{{I}_{2}}=2.0A,\,{{V}_{2}}=440\,V\] \[{{I}_{1}}=\frac{{{V}_{2}}{{I}_{2}}}{{{V}_{1}}}\times \frac{10}{8}=\frac{440\times 2\times 10}{220\times 8}=5A\]

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