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Question

Physics Question on Alternating current

When an ACA C source of voltage V=V0sin100tV=V_{0} \sin 100 t is connected across a circuit, the phase difference between the voltage VV and current II in the circuit is observed to be π/4\pi / 4, as shown in figure. If the circuit consists possibly only of RCR C or RLR L or LCL C in series, find possible values of two elements.

A

R=1kΩ,C=10μFR=1\, k\, \Omega,\, C=10\, \mu F

B

R=1kΩ,C=1μFR=1\, k\, \Omega,\, C=1\, \mu F

C

R=1kΩ,L=10mHR=1\, k\, \Omega,\, L=10\,m H

D

R=10kΩ,L=10mHR=10\, k\, \Omega,\, L=10\, m H

Answer

R=1kΩ,C=10μFR=1\, k\, \Omega,\, C=10\, \mu F

Explanation

Solution

Figure given in the question shows that current II leads the voltage VV by a phase angle π/4\pi / 4.
Therefore, the circuit can be RCR C circuit alone.
tanϕ=XCR=1ωCR\tan \phi=\frac{X_{C}}{R}=\frac{1}{\omega C R}
(XC=1ωC)\left(\because X_{C}=\frac{1}{\omega C}\right)
tanπ4=1ωCR;1=1ωCR\tan \frac{\pi}{4}=\frac{1}{\omega C R} ; 1=\frac{1}{\omega C R} ...(i)
From V=V0sin100tV=V_{0} \sin 100\, t,
we get, ω=100rads1\omega=100\, rad s^{-1}
CR=1ω=1100\therefore C R=\frac{1}{\omega}=\frac{1}{100} (using (i))
When R=1kΩ=103Ω,R=1\, k\, \Omega=10^{3} \Omega,
C=1105=105F=10μFC=\frac{1}{10^{5}}=10^{-5} F=10\, \mu F