Two resistors, R1 = 2.81Ω and R2 = 5.97Ω , are connected in series to a battery with an EMF of 24.0V and negligible internal resistance. Find the current I1 through R1 and the potential difference V2 across R2 .

2.73 A,16.3 V

Explanation:

When internal resistance is negligible

Then, V=E=24 V

In series

Using the formula

In series , current flowing through each resistance remains same and potential across each resistor is different.

Therefore,

f=ma newton's second law 7x8=56n

if they have a finite resultant