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

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

(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

(c) The Potential difference across the second resistor is

V₂ = IR₂ Equation 4

V₂ = 0.273×6

V₂ = 1.638 V