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When an A.C. source of e.m.f. \mathrm{E=E_0 \sin (100 t)} is connected across a circuit, the phase difference between the e.m.f. E and current I is observed to be \mathrm{\pi / 4} as shown in figure. If the circuit consists possibly only of RC or RL or L – C in series, the relationship between the two elements is
                                                            

Option: 1

\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=10 \mu \mathrm{F}


Option: 2

\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=1 \mu \mathrm{F}


Option: 3

\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{L}=10 \mathrm{H}


Option: 4

\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{L}=1 \mathrm{H}


Answers (1)

best_answer

\mathrm{Current\, \, leads \, \, by\, \pi / 4 \, \, and \, \, \omega=100 }
\mathrm{Now \tan \phi=\frac{1}{\omega \mathrm{RC}} or \tan \frac{\pi}{4}=\frac{1}{100 \mathrm{RC}} }
\mathrm{1=\frac{1}{100 \mathrm{RC}} or \mathrm{RC}=\frac{1}{100}=10^{-2} }
For combination (a)
\mathrm{\left.\mathrm{RC}=\left(1 \times 10^3\right) \times(10 \times 10)^{-6}\right)=10^{-2} }

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sudhir kumar

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