Question

A battery of internal resistance 4Ω is connected to the network of resistances as shown in fig, in order that maximum power can be delivered to the network, the value of R in ohm should be. 

(A) (4/9) 

(B) 2 

(C) (8/3) 

(D) 18

Answer 1

The correct option (B) 2 

Explanation:

It can be seen that the resistive network can be drawn as

Hence it is balanced wheat stones bridge ∴ No current through 6R.

hence Reff = (R + 2R) || (4R + 2R) i.e. 3R || 6R

∴ Reff = {(3R × 6R)/(3R + 6R)} = 2R

also I Total = {E/(Reff + R battery)} = {E/(4 + 2R)}

P = I_Total² (2R) = {E/(4 + 2R)}² × 2R

∴ (dP/dR) = 2E² (d/dR) [R/{4 + 2R}²]

(dP/dR) = 2E²[1(4 + 2R)^–2 – 2(4 + 2R)^–3 ∙ 2R]

for P max (dP/dR) = 0 hence (4 + 2R)^–2 = 4R (4 + 2R)^–3

∴ 1 = {(4R)/(4 + 2R)}

hence 4 + 2R = 4R

R = 2Ω