91Ó°ÊÓ

Emf of battery is the voltage difference across resistor network.

The power dissipated in equivalent resistance is the power dissipated in the circuit.

The expression for the power can be given as,

$\mathit{P}\mathbf{=}\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{R}}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(1)

Emf = 48.0 V

R₁ = R₂ = 4.00 Ω

R₄= 3.00 Ω

P = 295 W

Now, use the equation (1) to estimate the value of resistance such that,

$$\begin{gathered} P=\frac{V^2}{R} \\ 295W=\frac{{\left(48.0V\right)}^2}{R} \end{gathered}$$

Rearranging the above equation gives,

$$\begin{gathered} R=\frac{{\left(48.0V\right)}^2}{295W} \\ R=7.81\mathrm{Î\copyright } \end{gathered}$$

Which are the values of equivalent resistance.

Equivalent resistance from circuit resistance is,

The equivalent resistance of R₁ and R₂ which are in series

R₁₂ = R₁+R₂ = 4.00 Ω + 4.00 Ω = 8.00 Ω

The equivalent resistance of R₁₂ and R₂₃ which are in parallel, therefore

$\frac{1}{R_{123}}=\frac{1}{R_{12}}+\frac{1}{R_3}$ (2)

$R_{123}$and R₄ are in series thus,

$R=R_{123}+R_4$

Substitute all the values in the above expression, and we get,

$$\begin{gathered} R=R_{123}+R_4 \\ 7.81\mathrm{Î\copyright }=R_{123}+3.00\mathrm{Î\copyright } \\ {R}_{123}=7.81\mathrm{Î\copyright }=3.00\mathrm{Î\copyright } \\ {R}_{123}=4.81\mathrm{Î\copyright } \end{gathered}$$

Substitute the values of$R_{123}$and$R_{12}$in equation 2,

$$\begin{gathered} \frac{1}{4.81\mathrm{Î\copyright }}=\frac{1}{8.00\mathrm{Î\copyright }}+\frac{1}{R^3} \\ \frac{1}{R^3}=\frac{1}{4.81\mathrm{Î\copyright }}-\frac{1}{8.00\mathrm{Î\copyright }} \\ \frac{1}{R^3}=\frac{1}{12.1\mathrm{Î\copyright }} \\ {R}_3=12.1\mathrm{Î\copyright } \end{gathered}$$

Thus, the resistance $R_3$ for the resistor network to dissipate electrical energy is $12.1\mathrm{Î\copyright }$