Chapter 19, Problem 4P

To determine

The internal resistance of the cell.

Answer to Problem 4P

The internal resistance of the cell is $0.068\Omega$ .

Explanation of Solution

Given information:

Voltage of the dry cell, $v=1.5\text{V}$

Current across ammeter, $I=22\text{A}$

Formula used:

Write the expression to calculate the internal resistance of the cell using Ohm’s law.

  $v=Ir$............ (1)

Where,

• The internal resistance of the cell is $r$ .

Calculation:

Since, the internal resistance of the cell is much greater than the internal resistance of the ammeter. So, neglect the internal resistance of the ammeter and apply Ohm’s law.

Substitute the values in equation (1).

  $\begin{array}{c}1.5=22\times r\\ r=\frac{1.5}{22}\\ =0.068\Omega \end{array}$

Therefore, the internal resistance of the cell is $0.068\Omega$ .