Question

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 .

Answer 1

2.73 A,16.3 V

Step-by-step explanation:

`R_1=2.81\Omega`

`R_2=5.97\Omega`

`E=24 V`

When internal resistance is negligible

Then, V=E=24 V

In series

`R=R_1+R_2`

Using the formula

`R=2.81+5.97=8.78\Omega`

`V=IR`

`24=I(8.78)`

`I=(24)/(8.78)=2.73 A`

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

Therefore,
`I=I_1=I_2=2.73 A`

`V_2=IR_2`

`V_2=2.73* 5.97=16.3 V`