Three cells each of e.m.f. 1.5v and an internal resistance of 1.0\(\Omega\) are connected in parallel across a load resistance of 2.67\(\Omega\). Calculate the current in the load?

Destinedtowin

4 years ago

how do u get the effective internal resistance
not really clear enough

fabamz01

3 months ago

It said "3" cells of an internal resistance of 1 ohms was connected in parallel.... That would be


1/r=1/1+ 1/1+1/1 remember it is parallel, not series
nd that would be 1/r=3/1
r=1/3
r=0.33

1.5=I(0.33+ 2.67)
I= 1.5/3

I=0.5

Impulse12

2 years ago

how is the internal resistance gotten men i need a detailed explanation

Micahlistic

11 months ago

it is D. not A.

jonathanolatunji644

3 years ago

I think the addition of the three parallel resistance should be equal to one,
(1/r + 1/r + 1/r); 1/1 + 1/1 + 1/1 = 1