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## A 16 Ω resistor and a 6 Ω resistor are connected in series to an ideal 6 V battery. Find the current in each resistor. Answer in

Question

A 16 Ω resistor and a 6 Ω resistor are connected in series to an ideal 6 V battery.

Find the current in each resistor.

Answer in units of A.

b) Find the potential difference across the first

resistor.

Answer in units of V.

c) Find the potential difference across the second

resistor.

Answer in units of V

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Physics
3 years
2021-08-27T04:30:13+00:00
2021-08-27T04:30:13+00:00 2 Answers
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## Answers ( )

Answer:Explanation:a) series resistors carry the same current

A = V/Re = 6/(16 + 6) = 0.2727272… ≈ 27 mA

b) V = V₀(R/Re) = 6(16/(16 + 6)) = 4.363636 ≈ 4.4 V

c) V = V₀(R/Re) = 6(6/(16 + 6)) = 1.636363 ≈ 1.6 V

or V = 6 – 4.4 = 1.6 V

Answer:(a) 0.273 A(b) 4.368 V(c) 1.638 VExplanation:From the question,

(a) Applying ohm’s law

V = IR’…………………. Equation 1

Where V = Voltage of the battery, I = Current in each of the resistor, R’ = Total resistance of the combined resistors

Since the Two resisstors are connected in series,

(i) The same current flows through both resistors

(ii) The total resistor (R’) = R₁+R₂

Therefore,

V = (R₁+R₂)I

Make I the subject of the equation

I = V/(R₁+R₂)…………….. Equation 2

Given: V = 6 V, R₁ = 16 Ω, R₂ = 6 Ω

Substitute into equation 2

I = 6/(16+6)

I = 6/22

I = 0.273 A

(b) The potential difference across the first resisto

V₁ = IR₁…………………. Equation 3

Given: I = 0.273 A, R₁ = 16 Ω

Substitute these values into equation 3

V₁ = 0.273(16)

V₁ = 4.368 V

(c) The Potential difference across the second resistor is

V₂ = IR₂……………….. Equation 4

V₂ = 0.273×6

V₂ = 1.638 V