A cell of internal resistance 2Ω supplies current to a 6Ω resistor. The efficiency of the cell is

Richardq

7 years ago

for efficiency to be possible one needs to sum up both the internal resistance i.e (R+r) then the cumulative should be used to divide the external resistance and multiply by 100
(i.e efficiency= R/(R+r)*100.) that is it should be. efficiency=6/(6+2)*100 the right answer is 75%

Ojosamuel

8 years ago

The formula to use should be R/R+r × 100% not r/R × 100%
please reconsider, and feed me back
Thanks.

olyezema

4 years ago

i think this is the formula

Emmanueloluwaseun

5 years ago

myschool pls the answer here is 75%

R/r+R ×100

6/8×100=75%

Okewale2007

3 years ago

R/R+r *100 is the formular which is 75

okraDEN2023

2 years ago

The answer is D.

Efficiency = IR/I(R+r) x 100 The current cancels out. Then

Efficiency = R/(R+r) x 100
= 6/(6+2) x 100
= 6/8 × 100
= 75%

Stumer

2 years ago

The answer in the video did not correspond to the explanation answer already given

Sensei

2 months ago

75% correct 💯

Joywwe

2 years ago

Please the answer is 75% not 33.3% u can change this