Question 124: Current Electricity

Question

A battery with EMF $E$ and internal resistance $r$ is connected across a resistance $R$. The power consumption in $R$ will be maximum when:

Accepted Answer

The total resistance of the circuit is $(R+r)$.
Using Ohm's Law, the current I flowing through the circuit is:
$$ I=\frac{E}{R+r} $$
The power consumption P in the external resistor R is given by the formula:
$$ \mathrm{P}=\mathrm{I}^2 \mathrm{R} $$
Substituting the expression for I:
$$ P=\left(\frac{E}{R+r}\right)^2 R=\frac{E^2 R}{(R+r)^2} $$
To find the value of R for which P is maximum, we differentiate P with respect to R and set the derivative to zero ( $\frac{\mathrm{dP}}{\mathrm{dR}}=0$ ).
$$ \frac{d P}{d R}=E^2\left[\frac{(R+r)^2 1-R \cdot 2(R+r)}{(R+r)^4}\right]=0 $$
$$\Rightarrow $$ $(R+r)^2-2 R(R+r)=0$
$$\Rightarrow $$ $(R+r)-2 R=0$
$$\Rightarrow $$ $r-R=0$
$$\Rightarrow $$ $$ \mathrm{R}=\mathrm{r} $$
Therefore, the power consumption in the external load R is at its maximum when the external resistance is equal to the internal resistance of the battery, $\mathrm{r}=\mathrm{R}$.
Hence, the correct option is (D).

Answer

$R = \sqrt{2} \, r$

Answer

$R = 2r$

Answer

$R = \frac{r}{2}$

Answer

$R = r$

Current Electricity

A battery with EMF E and internal resistance r is connected across a resistance R. The power consumption in R will be maximum when:

Choose the correct answer: