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Question
Two resistors of 2.0 Ω and 3.0 Ω are connected (a) in series (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.
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Solution
(a) In Series -
R1 = 2 Ω
R2 = 3 Ω
R = R1 + R2 = 2 + 3 = 5 Ω
V = 6 V
`"I" = "V"/"R"`
I `= 6/5`
I = 1.2 A
(b) In Parallel -
R1 and R2 are connected in parallel
`1/"R" = 1/"R"_1 + 1/"R"_2`
`1/"R" = 1/2+1/3`
`1/"R" = 5/6`
R = `6/5` = 1.2 Ω
V = 6V
I `= "V"/"R"`
`= 6/1.2`
= 5 A
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