A cell of internal resistance r supplies current to a 6.0Ω resistor and its efficiency is 75%. Find the value of r.
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 below.
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}