91Ó°ÊÓ

Let R₁ = 45.0 Ω, R₂ = 18.0 Ω and R₃ = 15.0 Ω. The internal resistance of the battery is r=3.26 Ω

The two resistors R₁ and R₃ are in parallel and calculating their combination by resistors in a parallel equation as:

${\mathit{R}}_{\mathbf{13}}\mathbf{=}\frac{{\mathbf{R}}_{\mathbf{1}}{\mathbf{R}}_{\mathbf{3}}}{{\mathbf{R}}_{\mathbf{1}}\mathbf{+}{\mathbf{R}}_{\mathbf{3}}}$

Substitute values to get R₁₃ as:

$$\begin{gathered} R_{13}=\frac{R_1{R}_3}{R_1+R_3} \\ =\frac{\left(45.0\mathrm{Î\copyright }\right)\left(15.0\mathrm{Î\copyright }\right)}{45.0\mathrm{Î\copyright }+15.0\mathrm{Î\copyright }} \\ =11.25\mathrm{Î\copyright } \end{gathered}$$

R₁₃ is in series with R₂ so the current is the same for both of them which also equals the current flows in all circuits. The battery has an internal resistance and the emf of the battery isε= 25.0 Vso the current flows in the circuit could be calculated by Ohm's law in the form as:

$\mathbf{l}\mathbf{=}\frac{\mathbf{Î\mu }}{{\mathbf{R}}_{\mathbf{t}}}$

where R_t is the total resistance in the circuit = R₁₃+ R₂+ r.

Substitute values to get the current that reads by the ammeter

$$\begin{gathered} \mathrm{l}=\frac{\mathrm{Î\mu }}{{\mathrm{R}}_{13}+{\mathrm{R}}_2+\mathrm{r}} \\ =\frac{\mathrm{Î\mu }}{11.25\mathrm{Î\copyright }+18.0\mathrm{Î\copyright }+3.26\mathrm{Î\copyright }} \\ =0.796\mathrm{A} \end{gathered}$$

Therefore, the reading of the ammeter is 0.769 A.