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`

Explanation

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 times 10Js^-1'4.2 times 10=J_1^2R`
or `I_1=sqrt((4.2 times 10)/5)=sqrt8.4=2.9A`
Since AB and CD are in parallel, the potential difference remains the same between C and D, and between A and B.
`therefore (I-I_1)(4+6)=I_1 times 5`, On solving `I_1` from Eq (i)
We get `(I-2.9)10=2.9 times 5 or I-2.9=1.45 or I=4.35`
Heat released per second in `4Omega` will be `(4.35-2.9)^2 times 4=8.4J//s=2cals^-1`