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Question
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. How much current would flow through the 12 Ω resistor?
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Solution
Given: R1 = 0.2 Ω
R2 = 0.3 Ω
R3 = 0.4 Ω
R4 = 0.5 Ω
R5 = 12 Ω
V = 9 V
Therefore, the resultant resistance is given as:
R = R1 + R2 + R3 + R4 + R5
R = 0.2 + 0.3 + 0.4 + 0.5 + 12
R = 13.4 Ω
The current flowing through the 12 Ω resistance is given as:
I = `V/R`
I = `9/13.4`
I = 0.67 ampere
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