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Q93P

Expert-verifiedFound in: Page 801

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Thermal energy is to be generated in a**** ${\mathbf{0}}{\mathbf{.10}}{\mathbf{}}{\mathbf{\text{\hspace{0.17em}\Omega}}}$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 $

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,$\epsilon =1.5\text{\hspace{0.17em}V}$ .

**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.

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.$1.0\text{\hspace{0.17em}V}$

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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