In this circuit , when certain current flows , the heat produced in `5 Omega` is 4.05 J in a time t . The heat produced in `2 Omega` coil in the same time interval is

Options

A. `5.76`
B. 1.44
C. 2.88
D. 2.02

Answer

C

Explanation

As per question ,
Heat produced in `5 Omega` in time t is 4.05 = `I_(1)^(2)` 5
`I_(1) = sqrt((4.05)/(5)) = 0.9 A`
Potential difference across B and C is = `5 I_(1) = 5 xx 0.9 V = 4.5 V `
`I_(2) = (4.5 V)/(15 Omega) = 0.3 A`
`therefore I = I_(1) +I_(2) = 0.9 A + 0.3 A = 1.2 A`
Heat produced in `2 Omega` in same time interval t is `= (1.2A)^(2) (2 Omega) = 2.88` J