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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, respectively. How much current would flow through the 12 Ω resistor?
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Solution
When equivalent resistances are connected in series,
R = R1 + R2 + R3 + R4 + R5
R1 = 0.2 Ω, R2 = 0.3 Ω, R3 = 0.4 Ω, R4 = 0.5 Ω, R5 = 12 Ω ....(Given)
R = 0.2 + 0.3 + 0.4 + 0.5 + 12
R = 13.4 Ω
V = IR ....(According to Ohm’s law)
I = `"V"/"R"`
= `9/13.4`
= 0.67 A
There will be no division of electric current because the resistors are connected in series.
0.67 A of electric current will flow in a 12 Ω resistor.
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