A battery with EMF E and internal resistance | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The current $I$ in the circuit is given by $I = \frac{E}{R+r}$.The power $P$ consumed by the external resistance $R$ is $P = I^2 R = \left(\frac{E

Vedclass · Accepted Answer

$R=r$

Vedclass · Answer

(D) The current $I$ in the circuit is given by $I = \frac{E}{R+r}$.The power $P$ consumed by the external resistance $R$ is $P = I^2 R = \left(\frac{E}{R+r}\right)^2 R$.To find the condition for maximum power, we differentiate $P$ with respect to $R$ and set it to zero: $\frac{dP}{dR} = 0$.$\frac{dP}{dR} = E^2 \left[ \frac{(R+r)^2(1) - R(2)(R+r)}{(R+r)^4} \right] = 0$.This implies $(R+r)^2 - 2R(R+r) = 0$.Dividing by $(R+r)$, we get $(R+r) - 2R = 0$, which simplifies to $r - R = 0$ or $R = r$.Thus, the power consumption is maximum when the external resistance equals the internal resistance of the battery.

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

Similar Questions

An amount of charge $Q$ passes through a coil of resistance $R$. If the current in the coil decreases to zero at a uniform rate during time $T$, then the amount of heat generated in the coil will be,

$A$ time-varying current $i$ is passed through a resistance $R$ as shown in the figure. The total heat generated in the resistance is

$A$ household supply is $220 \, V$ with a fuse of $9 \, A$. What is the maximum number of $60 \, W$ bulbs that can be connected in parallel and operated simultaneously?

$A$ current of $2\, A$ passing through a conductor produces $80\, J$ of heat in $10$ seconds. The resistance of the conductor is ............ $\Omega$.

If ${R_1}$ and ${R_2}$ are respectively the filament resistances of a $200 \ W$ bulb and $100 \ W$ bulb designed to operate on the same voltage,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module