In the circuit shown in Fig. `7.39`, the heat produced in the `5 Omega` resistor due to the current flowing through it is `10 cals^(-1)` . The heat generated in the `4 Omega` resistor is

Options

A. `1cals^(-1)`
B. `2cals^(-1)`
C. `3cals^(-1)`
D. `4cals^(-1)`

Answer

B

Explanation

b. Let `I_1` be the current flowing in `5Omega` and `(I-I_1)"in" 4Omega` and 6Omega`.

The heat generated in the `5Omega` resistor is `10cals^(-1) = 4.2 xx 10Js^(-1)`.
`4.2 xx 10 = I_(1)^(2) R `
or `I_1 = sqrt((4.2 xx 10)/(5)) = sqrt(8.4) = 2.9A`
Since AB and CD are in parallel, the potential difference
remains the same between C and D, and between A and B.
`:. (I-I_1)(4+6) = I_1 xx 5`
On solving using `I_1` from Eq. (i),
we get,
`(I-2.9)10 = 2.9 xx 5`
or `I- 2.9 = 1.45`
or `I= 4.35`
Heat released per second in `4Omega` will be `(4.35 - 2.9)^2 xx 4 = A2.4`
`Js^(-1) = 2cals^(-1).`