91Ó°ÊÓ

According to Ohm’s Law, the current can be calculated if the resistance R of the circuit and V voltage drop across it given, as:

$\mathit{I}\mathbf{=}\frac{\mathbf{V}}{\mathbf{R}}$

Ammeter reads I = 1.25 A, therefore the current through the 25 Ω resistor is I₂₅= 1.25 A, and the voltage across the resistor is:

$$\begin{gathered} V_{ab}=\left(1.25A\right)\left(25\mathrm{Î\copyright }\right) \\ =31.25V \end{gathered}$$

The other resistors in the parallel circuit also have the same voltage

So, the current through the 15 Ω resistor is :

$$\begin{gathered} l_{15}=\frac{31.25}{15\mathrm{Î\copyright }} \\ =2.08A \end{gathered}$$

the current through (15+10) Ω resistor is

$$\begin{gathered} l_{\left(15+10\right)}=\frac{31.25}{25\mathrm{Î\copyright }} \\ =1.25A \end{gathered}$$

The sum of the above three currents is the total current in the circuit

$$\begin{gathered} l_{ab}=l_{25}+l_{15}+l_{\left(15+10\right)} \\ =1.25A+2.08A+1.25A \\ =4.58A \end{gathered}$$

Since there are no other branches, I_ab flows through 45Ω and 35 Ω. Thus, the voltage across 45 Ω is given by

$$\begin{gathered} V_{45}=\left(4.58A\right)\left(45\mathrm{Î\copyright }\right) \\ =206.10V \end{gathered}$$

Therefore, the voltmeter reads 206.1 V.

Since the battery has negligible resistance, therefore the emf of the battery will equal the three voltage as next

$$\begin{gathered} \mathrm{Î\mu }=V_{ab}+V_{45}+V_{35} \\ =31.25V+2.06.10 V+\left(4.58A\right)\left(35\mathrm{Î\copyright }\right) \\ =397.65V \end{gathered}$$

Thus, the emf of E of the battery is 397.65 V