Question

a battery of emf E and negligible resistance is connected to 2 resistancs R1 and R2 in series what is the potential difference across R2

Answer 1

1. Find the current through the circuit first
2. this will also be the curent through R2
3. potential difference across R2 = curent through R2*R2

Answer 2

lnternal resistance of the electrical source is is r, while the resistance of the lamp is R.so check out this: R/R+r
lls this correct

Answer 3

sorry ignore my previous post
I meanE/(R1+R2)/R2

Answer 4

yes,
E/(R1+R2)/R2 can also be written as (E*R2)/(R1+R2)

Answer 5

ok thanks

Answer 6

you're welcome