Question

An idealized voltmeter is connected across the terminals of a$\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{-}\mathit{V}$battery, and arole="math" localid="1655719696009" $\mathbf{75}\mathbf{.}\mathbf{0}\mathbf{-}\mathit{\Omega }$ appliance is also connected across its terminals. If the voltmeter reads$\mathbf{11}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{V}$ (a) how much power is being dissipated by the appliance, and (b) what is the internal resistance of the battery?

Short answer

(a) The power dissipated by the appliance is$1.89\mathrm{W}$.

(b) The internal resistance of the battery is$19.62\Omega$.

Step-by-step solution

Define the ohm’s law, resistance (R)and power (P).

According to Ohm’s law, the current flowing through the conductor is directly proportional to the voltage across the two points.

$V=IR$

Where,$I$is current in ampere $\left(A\right)$, is resistance in ohms $\left(\Omega \right)$and $V$is the potential difference volt $\left(V\right)$.

If real source of emf $\epsilon$has internal energy $r$then its terminal potential difference $V_{ab}$depends upon current.

So, formula for the terminal potential difference is:

$V_{ab}=\epsilon -Ir$

The ratio of V to$\mathit{I}$ for a particular conductor is called its resistance$\mathit{R}$

$R=\frac{V}{I}\mathrm{or}\frac{pL}{A}$

Where, $p$is resistivity$\Omega ‐\mathrm{m},L$is length in $\mathrm{m}$and $A$is area in ${\mathrm{m}}^2$.

The power $\left(P\right)$is the product of potential difference $\left(V\right)$and the current$\left(I\right)$ .

$P=VI\mathrm{or}{I}^{2}Ro\mathrm{r}\frac{{\mathrm{V}}^2}{\mathrm{R}}$

Determine the power.

Given that,

$$\begin{gathered} V=11.9V \\ R=75\Omega \end{gathered}$$

The power dissipated in the circuit is:

$$\begin{gathered} P=\frac{V^2}{R} \\ =\frac{11.9^2}{75} \\ =1.89\mathrm{W} \end{gathered}$$

Hence, the power dissipated by the appliance is

$1.89\mathrm{W}$

Determine the internal resistance.

Given that,$\epsilon =15.0V$

The internal resistance of the battery is:

$$\begin{gathered} r=\frac{\epsilon -V}{I} \\ \Rightarrow r=\frac{\epsilon -V}{\left({\displaystyle \frac{V}{R}}\right)} \end{gathered}$$

Substitute the values of variables

$$\begin{gathered} r=\frac{15-11.9}{\left({\displaystyle \frac{11.9}{75}}\right)} \\ =\frac{3.1}{0.158} \\ =19.62\Omega \end{gathered}$$

Hence, the internal resistance of the battery is$19.62\Omega$.