Two resistors, R1=2.79 Ω and R2=6.37 Ω , are connected in series to a battery with an EMF of 24.0 V and negligible internal resistance. Find the current I1 through R1 and the potential difference V2 across R2 .

Current Through R₁ = I₁ = 2.62 A and V₂ = 16.6 V

Explanation:

To find I₁ across R₁

For series circuit same amount of current are flowing through all resister ie

I₁ = I₂ = I

To find I₁ we have to calculate R(eq)=R₁ + R₂ =2.79Ω + 6.37Ω = 9.16 Ω

V = 24.0 V (given)

So I = I₁ = V / R(eq) = 24 V / 9.16 Ω =2.62 A

To find V₂ across R₂

As in series circuit the potential difference across each resisters are of different amount depending upon the resistance of resister,  so V₂ = I R₂

⇒ V₂ = 2.62 A × 6.37 Ω = 16.6 V

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