Question

A battery of internal resistance $4\,\Omega$ is connected to the network of resistance as shown. In the order that the maximum power can be delivered to the network, the value of $R$ in ohm should be

A. $\dfrac{4}{9}$
B. $2$
C. $\dfrac{8}{3}$
D. $18$

Answer 1

As, the given circuit is complicated so we first need to simplify the circuit. Then find the equivalent resistance of the circuit. We are asked to find the value of R when maximum power is delivered to the circuit. Recall the condition for which maximum power is delivered to the circuit.

Complete step by step answer:
Given, internal resistance of battery $r = 4\,\Omega$
We would need to find the equivalent resistance of the circuit, for that, first let’s name the branches.

We observe that resistance of branch AF and FE are in series so equivalent resistance of branch AE will be,

\Rightarrow {R_{AE}} = R + R = 2R $$ Similarly, we observe that resistance of branch BC and CD are also in series so equivalent resistance of branch BD will be, $${R_{BD}} = {R_{BC}} + {R_{CD}} \\\ \Rightarrow {R_{BD}} = R + R = 2R $$ Now, we draw a simplified circuit with $${R_{BD}}$$ and $${R_{AE}}$$  We observe that the circuit ABDEA forms a Wheatstone bridge as the ratio of resistance on branch AB and AE is equal to the ratio of resistance on branch BD and DE. That is, $$\dfrac{{{R_{AB}}}}{{{R_{AE}}}} = \dfrac{R}{{2R}} \\\ \Rightarrow\dfrac{{{R_{AB}}}}{{{R_{AE}}}} = \dfrac{1}{2}$$ and $$\dfrac{{{R_{BD}}}}{{{R_{DE}}}} = \dfrac{{2R}}{{4R}}\\\ \Rightarrow\dfrac{{{R_{BD}}}}{{{R_{DE}}}} = \dfrac{1}{2}$$ $$ \Rightarrow \dfrac{{{R_{AB}}}}{{{R_{AE}}}} = \dfrac{{{R_{BD}}}}{{{R_{DE}}}}$$ As ABDEA forms a Wheatstone bridge so, no current will flow through the branch BE and we can eliminate this branch. Now, we draw a more simplified circuit.  Now, we find the equivalent resistance on branch ABD, as the resistances are in series, so $${R_{ABD}} = 2R + 4R = 6R$$ Similarly for equivalent resistance on branch AED, the resistances are in series, so $${R_{AED}} = R + 2R = 3R$$ Now, we find the equivalent resistance of the whole circuit, since the resistance $${R_{AED}}$$ and $${R_{ABD}}$$ are in parallel, so equivalent resistance of the circuit will be $$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_{ABD}}}} + \dfrac{1}{{{R_{AED}}}}$$ Putting the values of $${R_{AED}}$$ and $${R_{ABD}}$$, we get $$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{6R}} + \dfrac{1}{{3R}} \\\ \Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{3}{{6R}} \\\ \Rightarrow {R_{eq}} = 2\,R\,\Omega $$ For maximum power to be delivered to the network, equivalent resistance must be equal to in internal resistance. That is, $${R_{eq}} = r$$ Now, we put the value of internal resistance $$2R = 4 \\\ \therefore R = 2\,\Omega $$ **Hence, the correct answer is option B.** **Note:** Whenever there is a complex circuit, always try to simplify the circuit. Always remember the formulas for resistance in parallel and for resistance in series and check carefully before using it. And remember the condition for Wheatstone bridge can be used only when the ratios of the resistance of both the arms of the circuit are equal.