Question
A battery of internal resistance $$4\Omega $$ is connected to the network of resistances as shown. In order that the maximum power can be delivered to the network, the value of $$R$$ in $$\Omega $$ should be
A battery of internal resistance $$4\Omega $$ is connected to the network of resistances as shown. In order that the maximum power can be delivered to the network, the value of $$R$$ in $$\Omega $$ should be
A.
$$\frac{4}{9}$$
B.
2
C.
$$\frac{8}{3}$$
D.
18
Answer :
2
Solution :
The equivalent circuits are shown in the figure.
The circuit represents balanced Wheatstone Bridge. Hence $$6R\,\Omega $$ resistance is ineffective
$$\frac{1}{{{R_{eq}}}} = \frac{1}{{3R}} + \frac{1}{{6R}},\,\,\,{R_{eq}} = \frac{{\left( {3R} \right)\left( {6R} \right)}}{{\left( {3R} \right) + \left( {6R} \right)}} = 2R$$
For Max. Power
External Resistance = Internal Resistance
$$2R = 4\Omega \,\,\therefore R = 2\Omega $$
The equivalent circuits are shown in the figure.
The circuit represents balanced Wheatstone Bridge. Hence $$6R\,\Omega $$ resistance is ineffective
$$\frac{1}{{{R_{eq}}}} = \frac{1}{{3R}} + \frac{1}{{6R}},\,\,\,{R_{eq}} = \frac{{\left( {3R} \right)\left( {6R} \right)}}{{\left( {3R} \right) + \left( {6R} \right)}} = 2R$$
For Max. Power
External Resistance = Internal Resistance
$$2R = 4\Omega \,\,\therefore R = 2\Omega $$
