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Expert-verified Found in: Page 801 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Thermal energy is to be generated in a ${\mathbf{0}}{\mathbf{.10}}{\mathbf{}}{\mathbf{\text{\hspace{0.17em}Ω}}}$resistor at the rate of ${\mathbf{10}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbf{}}{\mathbf{\text{W}}}$by connecting the resistor to a battery whose emf is${\mathbf{1}}{\mathbf{.5}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbf{}}{\mathbf{\text{V}}}$. (a) What potential difference must exist across the resistor? (b) What must be the internal resistance of the battery?

a) The potential difference that must exist across the resistor is.$1.0\text{\hspace{0.17em}V}$

b) The internal resistance of the battery is.$0.05\text{\hspace{0.17em}}\Omega$

See the step by step solution

## Step 1: The given data

a) Resistance of the resistor,$R=0.10\text{\hspace{0.17em}}\Omega$ .

b) Rate of energy transferred as thermal energy,$P=10\text{\hspace{0.17em}W}$ ,

c) Emf of the battery,$\epsilon =1.5\text{\hspace{0.17em}V}$ .

## Step 2: Understanding the concept of energy rate transfer

Here, the rate of energy transferred to the resistor as thermal energy is given. Thus, using the concept, the value of the external resistance can be given. Now, the current through the battery and the resistor is equal to the closed loop rule. Thus, equating the current values will determine the internal resistance of the battery.

Formulae:

The rate at which the thermal energy is transferred,

$P=\frac{{V}^{2}}{R}$ (i)

The voltage equation using Ohm’s law,

$V=IR$ (ii)

Here $I$is the current, and $R$ is the resistance.

## Step 3: a) Calculation of the potential difference across the resistor

The potential difference across the resistor due to the rate of energy transferred as thermal energy can be given using equation (i) as follows:

$V=\sqrt{PR}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}V=\sqrt{\left(10\text{\hspace{0.17em}W}\right)\left(0.10\text{\hspace{0.17em}}\Omega \right)}\\ =1.0\text{\hspace{0.17em}V}\end{array}$

Hence, the value of the potential difference is.$1.0\text{\hspace{0.17em}V}$

## Step 4: b) Calculation of the internal resistance of the battery

Now, we know that current through the resistor and the battery is equal, and thus, the internal resistance of the battery can be given using equation (ii) as follows:

$\begin{array}{c}{i}_{R}={i}_{battery}\\ \frac{V}{R}=\frac{\epsilon -V}{r}\\ r=\left(\frac{\epsilon -V}{V}\right)R\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}r=\left(\frac{1.5\text{\hspace{0.17em}V}-1.0\text{\hspace{0.17em}V}}{1.0\text{\hspace{0.17em}V}}\right)0.10\text{\hspace{0.17em}}\Omega \\ =0.05\text{\hspace{0.17em}}\Omega \end{array}$

Hence, the value of internal resistance is.$0.05\text{\hspace{0.17em}}\Omega$

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