Suggested languages for you:

Americas

Europe

Q30PE

Expert-verified
Found in: Page 776

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (a) What is the internal resistance of a $$1.54{\rm{ }}V$$ dry cell that supplies $$1.00{\rm{ }}W$$ of power to a $$15.0{\rm{ }}\Omega$$ bulb?(b) What is unreasonable about this result?(c) Which assumptions are unreasonable or inconsistent?

a) Value of the internal resistance is $$r \approx - 9{\rm{ }}\Omega$$.

b) Internal resistance cannot be negative.

c) The value of power given in problem is not possible.

See the step by step solution

## Step 1: Concept Introduction

Internal resistance, which causes heat to be produced, is the opposition to the flow of current provided by the cells and batteries themselves. Ohms are used to measure internal resistance.

## Step 2: Given Information

• Emf of the cell:$$1.54 - V$$
• Power for thebulb:$$1.00 - W$$
• Current of the bulb:$$15.0 - \Omega$$

## Step 3: Calculation for the internal resistance.

a.

Calculate the internal resistance of a $$E = 1.54 - V$$dry cell that supplies a power of $$P = 1.00 - W$$ to a light bulb with resistance$$R = 15.0 - \Omega$$.

To do this, we first calculate the current that goes through the light bulb by using the equation for the power in a $$P = {I^2}R$$.

This gives

\begin{align}{}I & = \sqrt {\frac{P}{R}} \\I & = \sqrt {\frac{{1\;W}}{{15{\rm{ }}\Omega }}} \\I & = 0.258\;A\end{align}

The internal resistance is obtained by using the loop law, which gives

\begin{align}{}E - Ir - IR = 0\\r = \frac{E}{I} - R\\r & = \frac{{1.54\;V}}{{0.258\;A}} - 15{\rm{ }}\Omega \approx - 9{\rm{ }}\Omega \end{align}

Therefore, the internal resistance is$$- 9{\rm{ }}\Omega$$.

## Step 4: Calculation for the voltage of battery.

b)

The resistance can never be negative, so the former result is inconsistent.

## Step 5: Explanation

c)

A battery's voltage output is always limited by its emf. The voltage drop across $$R$$ should, in an ideal situation with no internal resistance, be equal to the emf, but if we calculate it using the current we computed, we get

\begin{align}{}V & = IR\\IR& = (0.258\;A) \times (15{\rm{ }}\Omega )\\IR & = 3.87\;V\end{align}

Which is greater than the battery's emf.

This indicates that the solution to the problem's power is illogical.