1.

The current (i_(1))(in A ) flowing through 1 Omega resistor in the following circuit is: (a) 0.50 (b) 0.30 (c) 0.25 (d) 0.20

Answer»

`0.50`
`0.30`
`0.25`
`0.20`

Solution :`0.20`

Equivalent resistance between C and D,
`R_(1) = (1xx 1)/(1 + 1) = (1)/(2) Omega `
Equivalent resistance between B and E,
`R_(2) = R_(1) + 2 = (1)/(2) = (1)/(2) + 2 = (5)/(2)`
Now, equivalent resistance of `R_(2) and 2 Omega` of AF,
`R_(3) =(R_(2) XX 2)/(R_(2) + 2) = ((5)/(2) xx 2 )/((5)/(2) + 2)= (5 xx 2)/(9) = (10)/(9) Omega`
In given circuit, voltage of BE = Voltage of AF = 1 V
`therefore ` Current flowing through BE= `(V)/(R_(2)) =(1)/((5)/(2)) = (2)/(5) `A
Now, current flowing in B-C-O-E is same and itis `(2)/(5)` A
`therefore `Current flowing through any resistorof `1 Omega`,
`I = (1)/(2) xx (2)/(5)`
`thereforeI = (1)/(5)"" therefore I = 0.20 `A


Discussion

No Comment Found

Related InterviewSolutions