A cell of internal resistance r supplies current to a 6.0Ω resistor and its efficiency…

Explanation

Efficiency = \(\frac{power output}{power input}\) = \(\frac{R}{R + r}\)
Given 

efficiency = 75% = 0.75, R = 6.0Ω, r = ?

Efficiency =  \(\frac{R}{R + r}\)

0.75 =  \(\frac{6.0}{6.0 + r}\)

0.75(6.0 + r) = 6.0

4.5 + 0.75r = 6.0

0.75r = 6.0 - 4.5 = 1.5

r =  \(\frac{1.5}{0.7}\) = 2Ω

The internal resistance r of the cell is 2.0 ohms.

There is an explanation video available .