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

Question

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

ok thanks

anonymous · Answer

you're welcome