Question

For the circuit shown in Fig. E26.6 both meters are idealized, the battery has no appreciable internal resistance, and the ammeter reads 1.25 A. (a) What does the voltmeter read?

(b) What is the emf E of the battery?

Short answer

(a) The voltmeter reads 206.1 V.

(b) The emf of E of the battery is 397.65 V

Step-by-step solution

Current

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:

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

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

So, the current through the 15 Ω resistor is :

$$\begin{gathered} l_{15}=\frac{31.25}{15\Omega } \\ =2.08A \end{gathered}$$

the current through (15+10) Ω resistor is

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

Calculation of the voltage

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

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

Therefore, the voltmeter reads 206.1 V.

Electromotive force of the battery

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

$$\begin{gathered} \epsilon =V_{ab}+V_{45}+V_{35} \\ =31.25V+2.06.10\hspace{0.17em}V+\left(4.58A\right)\left(35\Omega \right) \\ =397.65V \end{gathered}$$

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