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Question
A battery of e.m.f 16 V and internal resistance 2 Ω is connected to two resistors 3Ω and 6Ω connected in parallel. Find:
- the current through the battery.
- p.d. between the terminals of the battery,
- the current in 3 Ω resistors,
- the current in 6 Ω resistor.
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Solution
E.m.f. of battery E = 16V, r = 2Ω
Effective resistance of 3Ω and 6Ω connected in parallel R is
`1/"R" = 1/3 + 1/6`
= `3/6`
= `1/2`
∴ R = 2Ω
Total resistance = (R + r)
= (2 +2)
= 4Ω
(a) Current through battery
I = `"V"/("R" + r) = "E"/("R" +r)`
I = `(16V)/(4Ω)`
= 4 A
(b) p.d between terminal of battery
V = IR
= 4 × 2
= 8V
(c) The current in 3Ω resistor
I1 = `"V"/"R"`
= `8/3`
= 2.66 A
(d) The current in 6Ω resistor
I2 = `"V"/"R"`
= `8/6`
= 1.34 A
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