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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 .

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lublana

Answer:

2.73 A,16.3 V

Explanation:

[tex]R_1=2.81\Omega[/tex]

[tex]R_2=5.97\Omega[/tex]

[tex]E=24 V[/tex]

When internal resistance is negligible

Then, V=E=24 V

In series

[tex]R=R_1+R_2[/tex]

Using the formula

[tex]R=2.81+5.97=8.78\Omega[/tex]

[tex]V=IR[/tex]

[tex]24=I(8.78)[/tex]

[tex]I=\frac{24}{8.78}=2.73 A[/tex]

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

Therefore, [tex]I=I_1=I_2=2.73 A[/tex]

[tex]V_2=IR_2[/tex]

[tex]V_2=2.73\times 5.97=16.3 V[/tex]

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