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Answer :
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
(c) The Potential difference across the second resistor is
V₂ = IR₂.................... Equation 4
V₂ = 0.273×6
V₂ = 1.638 V
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