Chapter 19, Problem 2P

To determine

The internal resistance of four identical cells connected in series to a $12\Omega$ light bulb.

Answer to Problem 2P

The internal resistance of four identical cells connected in series to a $12\Omega$ light bulb is $0.33\Omega$ .

Explanation of Solution

Given information:

Voltage across each cell, $v=1.5\text{V}$

Resistance of the light bulb, $R=12\Omega$

Resulting current in the circuit, $I=0.45\text{A}$

Formula used:

Write the expression to calculate the current in the circuit using Kirchhoff Voltage Law.

  $-4v+4Ir+IR=0$............. (1)

Where,

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

Calculation:

Draw the circuit diagram.

  

Rearrange the equation (1) to calculate the internal resistance.

  $\begin{array}{c}-4v+4Ir+IR=0\\ 4Ir=4v-IR\\ r=\frac{4v-IR}{4I}\end{array}$

Substitute the values in the above equation.

  $\begin{array}{c}r=\frac{4\left(1.5\right)-\left(0.45\times 12\right)}{4\left(0.45\right)}\\ =\frac{6-5.4}{1.8}\\ =0.33\Omega \end{array}$

Therefore, the internal resistance of four identical cells connected in series to a $12\Omega$ light bulb is $0.33\Omega$ .