The maximum power drawn from a source with el | vedclass

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

Question

(B) The current $i$ in the circuit is given by $i = \frac{E}{r + R}$,where $R$ is the external resistance.Power $P$ delivered to the external resistan

Vedclass · Accepted Answer

$E^2 / 4r$

Vedclass · Answer

(B) The current $i$ in the circuit is given by $i = \frac{E}{r + R}$,where $R$ is the external resistance.Power $P$ delivered to the external resistance is $P = i^2 R = \frac{E^2 R}{(r + R)^2}$.To find the maximum power,we differentiate $P$ with respect to $R$ and set it to zero: $\frac{dP}{dR} = E^2 \left[ \frac{(r + R)^2 - R(2)(r + R)}{(r + R)^4} \right] = 0$.This simplifies to $(r + R) - 2R = 0$,which gives $R = r$.Substituting $R = r$ into the power equation: $P_{\max} = \frac{E^2 r}{(r + r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}$.

The maximum power drawn from a source with electromotive force $E$ and internal resistance $r$ is given by:

Similar Questions

If a power of $100\, W$ is being supplied across a potential difference of $200\, V$,the current flowing is ............ $A$.

An expression for the rate of heat generated,if a current of $I$ ampere flows through a resistance of $R$ $\Omega$,is

Why is electric power transmitted at very high voltage over long distances to reduce Ohmic loss?

$A$ Watt-hour meter measures:

Two identical electric lamps marked $500\, W, 220\, V$ are connected in series and then joined to a $110\, V$ line. The power consumed by each lamp is

Vedclass Test Series

Exam Paper Generator

Online Exam Module