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3 years ago

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

Physics

2 answers:

GrogVix [38]3 years ago

3 0

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

mojhsa [17]3 years ago

3 0

Answer:

(a) 0.273 A

(b) 4.368 V

(c) 1.638 V

Explanation:

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

V₂ = IR₂.................... Equation 4

V₂ = 0.273×6

V₂ = 1.638 V

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