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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 0.12 Ω. How much current would flow through the 12 Ω resistor?
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Solution
When five resistors are connected in series, the total resistance is given by:
R = R1+ R2+ R3+ R4+ R5
Here,
R1= 0.2 Ω
R2= 0.3 ,Ω
R3= 0.4 ,Ω
R4= 0.5 ,Ω
R5= 0.12,Ω
Therefore:
R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω
R = 13.4 Ω
Total resistance of the circuit = 13.4 Ω
The current flowing through this series combination is given by I = V / R.
or I = 9 / 13.4 = 0.67 A
Now since the resistors are connected in series, the current flowing through each resistance is the same. Hence, the current through the 12 Ω resistor is equal to 0.67 A.
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