Question

A battery of emf 10 V and internal resistance of 0.5 ohm is connected across a variable resistance R. The value of R for maximum power transfer is given by

• A. $0.5\Omega$
• B. $1.00\Omega$
• C. $2.0\Omega$
• D. $0.25\Omega$

Answer 1

The correct option is A $0.5\Omega$

$\text{Step 1: Draw circuit diagram and apply KVL in loop}$ $\text{[Ref. Fig. 1]}$

$E-iR-ir=0$

$i=\frac{E}{R+r}$

$\text{Step 2: Power dissipation in R}$ $\text{[Ref. Fig. 2]}$

$P=i^{2}R$

$P={\left(\frac{E}{R+r}\right)}^{2}R=\frac{E^{2}R}{(R+r{)}^2}$ $....\left(1\right)$

$\text{Step 3: Find R for maximum power dissipation}$

Given: $E=10V,r=0.5\Omega$

For maximum Power: $\frac{dP}{dR}=0,$ Therefore Using Equation $\left(1\right)$:

$\frac{dP}{dR}=\frac{(R+r{)}^2{E}^2-2E^{2}R(R+r)}{(R+r{)}^4}=0$

$\Rightarrow R+r=2R$

$\Rightarrow R=r$

$\text{Note:}$ We can also remember the above result that, for maximum power dissipation across external resistance, its value should be equal to the internal resistance of the battery.

$\Rightarrow R=0.5\Omega$

Hence, Option $\left(A\right)$ correct.