Question 69: Current Electricity

Question

A cell of emf 2 V is connected with a load of resistance 1.5 $$\Omega$$. The power delivered by the cell to the load is maximum, then power transferred to the load is

Accepted Answer

Given, emf of cell, $$E=2V$$
Load resistance, $$R_L=1.5~\Omega$$
Current flowing the circuit is
$$I=\frac{E}{R+r}$$
Power transferred to the load, R,
$$P = {I^2}R = {\left( {{E \over {R + r}}} ight)^2}\,.\,R = {{{E^2}R} \over {{{(R + r)}^2}}}$$
Power P will be maximum, if
$$dP/dR = 0$$
$$ \Rightarrow {d \over {dR}}{{{E^2}R} \over {{{(R + r)}^2}}} = 0$$
$$
\begin{array}{ll}
\Rightarrow & \frac{E^2(R+r)^2-E^2 \cdot R \cdot 2(R+r)}{(R+r)^3}=0 \
\Rightarrow & E^2(R+r)[R+r-2 R]=0 \
\Rightarrow & E^2(R+r)(r-R)=0 \
\Rightarrow & r-R=0 \
\Rightarrow & r=R
\end{array}$$
$$	herefore$$ Maximum power, $$P=\frac{E^2 R}{(R+r)^2}=\frac{E^2 R}{(R+R)^2}=\frac{E^2}{4 R}$$
$$\Rightarrow \quad P=\frac{2^2}{4 	imes 1.5}=\frac{1}{3}=0.33 \mathrm{~W}$$

Answer

0.33 W

Answer

2.67 W

Answer

1.33 W

Answer

3.25 W

Current Electricity

A cell of emf 2 V is connected with a load of resistance 1.5 $\Omega$. The power delivered by the cell to the load is maximum, then power transferred to the load is

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