91Ó°ÊÓ

Current$i=1mA$

Emf${\mathrm{Î\mu }}_2=3V$

Emf${\mathrm{Î\mu }}_1=2V$

Internal resistance$r_1=r_2=3Ω$

Use the Kirchhoff’s loop law to the given circuit and write an equation for R. Inserting the values in I will give us the value of resistorR and using the formula for the rate of the energy dissipation, calculate the thermal energy appeared in resistorR.

Kirchhoff's loop rule states that the sum of all the electric potential differences around a loop is zero.

Formulae are as follow

$$\begin{gathered} V=iR \\ P=i^{2}R \end{gathered}$$

Where,

i is current, V is voltage, R is resistance, P is power.

The resistance:

Applying Kirchhoff’s loop law,

${\mathrm{Î\mu }}_2-{\mathrm{Î\mu }}_1-i{r}_1-i{r}_2-iR=0$

Solving for resistance,

$({\mathrm{Î\mu }}_2-{\mathrm{Î\mu }}_1)-i(r_1+r_2)=iR$

$$\begin{gathered} R=\frac{(3V-2V)}{(1×{10}^{-3}A)}-(3Ω+3Ω) \\ R=994=9.9×{10}^2Ω \end{gathered}$$

Hence, the resistance R is$9.9×{10}^2Ω.$

Rate at which the thermal energy appears in R:

The rate of the dissipation of energy is given by,

$$\begin{gathered} P=i^{2}R \\ P=(1×{10}^{-3}A{)}^2(994Ω) \\ P=994×{10}^{-6}=9.9×{10}^{-4}W \end{gathered}$$

Hence, rate at which thermal energy appears in R is 9.9$×{10}^{-4}W$.

Therefore, use Kirchhoff’s law to find the value of resistance in the circuit and using the formula for the rate of energy dissipation, the thermal energy appeared in given resistance can be calculated.