Question

A$\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{-}\mathit{\Omega }$bulb is connected across the terminals of a$\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{V}$battery having$\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\Omega }$of internal resistance. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?

Short answer

$12.3\%$of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb.

Step-by-step solution

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

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

$V=IR$

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

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

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

Where,$p$is resistivity $\Omega -m,L$, is length in $m$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 $\left(I\right)$.

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

Determine the percentage of the power of the battery dissipated.

Given that,

$$\begin{gathered} R=12\Omega \\ \epsilon =12V \\ r=35\Omega \end{gathered}$$

The power dissipated by the internal energy is

$P_r=I^{2}r$

The total power consumed by the internal resistance and the bulb is

$P_t=I^{2}r\left(r+R\right)$

The ratio between the two power is

$$\begin{gathered} \frac{Pr}{Pt}=\frac{I^{2}r}{I^2\left(r+R\right)} \\ =\frac{r}{r+R} \\ =\frac{3.5}{3.5+25} \\ =12.3\% \end{gathered}$$

Hence,$12.3\%$ of the power of the battery is dissipated across the internal resistance.