दिये गये परिपथ में AC स्रोत का `omega = 100` rad/s है । प्रेरक तथा संधारित्र मानकर विकल्प का सही चुनाव करें ।

Options

A. a. परिपथ में धारा I = 0.3 A है ।
B. b. परिपथ में धारा I = 0.3 `sqrt(2) A` है
C. c. `100 Omega` प्रतिरोध के पार वोल्टेज `10 sqrt(2) V` है
D. d. `50 Omega` प्रतिरोध के पार वोल्टेज 10 V है ।

Answer

A, C

Explanation

`C = 100 mu F, (1)/(omega C) = (1)/(100 xx 100 xx 10^(-6)) = 10^(2)`
अतः `X_(C ) = (1)/(omega C) = 100 pi`
एवं `X_(L) = omega L = (100 (0.5) = 50 Omega`
अतः `Z_(1) = sqrt(X_(L)^(2) + (100)^(2)) = 100 sqrt(2) Omega`
एवं `Z_(2) = sqrt(X_(L)^(2) + (50^(2))`
`= sqrt((50^(2) + (50)^(2)) == 50 sqrt(2) Omega`
`e = 20 sqrt(2) sin omega t`

`I_(1) = (20 sqrt(2))/(100 sqrt(2)) sin (omega t + (pi)/(4))`
`= (1)/(5) sin (omega t + (pi)/(4))`
एवं `I_(2) = (20sqrt(2))/(50sqrt(2)) sin (omega t - (pi)/(4))`
अतः `I = sqrt(I_(1)^(2) + I_(2)^(2)) = sqrt((0.2)^(2) + (0.4)^(2))`
`= (0.2) sqrt(1 + 4) = (1)/(sqrt(5))`
`I_(max) = (1)/(sqrt(2) sqrt(5)) = (1)/(sqrt(10)) = (sqrt(10))/(10) = 0.2 A`
`(V_(100))_(max) = (I_(1))_(max) xx 100 = (0.2)/(sqrt(2)) xx 100`
`(V_(50))_(max) = (0.4)/(sqrt(2)) xx 50 = 10 sqrt(2) V`