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QJEEM19Ph33f0: JEE Main 2019 Physics - Class 12 JEEM19Ph33f0 Solution: In the above circuit, C = F, R 2 = 20 , L = H and R 1 = 10 .…

In the above circuit, C = 32{{\sqrt 3 } \over 2} μ\mu F, R2 = 20 Ω\Omega , L = 310{{\sqrt 3 } \over {10}} H and R1 = 10 Ω\Omega . Current in L-R1 path is I1 and in C-R2 path it is I2 . The voltage of A.C. source is given by, V = 200 2{\sqrt 2 } sin (100 t) volts . The phase difference between I1 and I2 is :

JEE Main 2019 (Online) 12th January Evening Slot Physics - Alternating Current Question 140 English

Multiple Choice Options

  • A 150o
  • B 90o
  • C 30o
  • D 0o

Step-by-Step Verified Solution

Phase difference between I2 and V, i.e. C −- R2 circuit is given by

tan⁡ϕ=XCR2⇒tan⁡ϕ=1CωR2\tan \phi = {{{X_C}} \over {{R_2}}} \Rightarrow \tan \phi = {1 \over {C\omega {R_2}}}

Substituting the given values, we get

tan⁡ϕ=132×10−6×100×20=1033\tan \phi = {1 \over {{{\sqrt 3 } \over 2} \times {{10}^{ - 6}} \times 100 \times 20}} = {{{{10}^3}} \over {\sqrt 3 }}

∴\therefore ϕ\phi 1 is nearly 90 ∘^\circ .

Phase difference between I1 and V, i.e. in L −- R1 circuit is given by

tan⁡ϕ2=−XLR1=−LωR\tan {\phi _2} = - {{{X_L}} \over {{R_1}}} = - {{L\omega } \over R}

Substituting the given values, we get

tan⁡ϕ2=−310×10010=−3\tan {\phi _2} = - {{{{\sqrt 3 } \over {10}} \times 100} \over {10}} = - \sqrt 3

As, tan⁡ϕ2=−3\tan {\phi _2} = - \sqrt 3

∴\therefore ϕ2=120∘{\phi _2} = 120^\circ

Now, phase difference between I1 and I2 is

Δϕ=ϕ2−ϕ1=120∘−90∘=30∘\Delta \phi = {\phi _2} - {\phi _1} = 120^\circ - 90^\circ = 30^\circ

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