There are two resistances of $4$ ohm and $12$ ohm. By connecting them with a cell of $10$ volt, how much will be total current in the circuit, if the resistances are connected
(i) in series (ii) in parallel?

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Solution

i) in Series
The equivalent resistance in series combination is,
$R_{e q} = R_{1} + R_{2} = 16 Ω$
$I = \frac{V}{R_{e q}}$
$I = \frac{10}{16} = 0.625 A$
II) In Parallel
The equivalent resistance in parallel combination is,
$R_{e q} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{4 \times 12}{12 + 4} = 3 Ω$
$I = \frac{V}{R_{e q}} = \frac{10}{3} = 3.33 A$

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There are two resistances of 4 ohm and 12 ohm. By connecting them with a cell of 10 volt, how much will be total current in the circuit, if the resistances are connected (i) in series (ii) in parallel?

i) in SeriesThe equivalent resistance in series combination is,Req=R1+R2=16 ΩI=VReqI=1016=0.625 AII) In ParallelThe equivalent resistance in parallel combination is,Req=R1R2R1+R2=4×1212+4=3 ΩI=VReq=103=3.33A

There are two resistances of $4$ ohm and $12$ ohm. By connecting them with a cell of $10$ volt, how much will be total current in the circuit, if the resistances are connected
(i) in series (ii) in parallel?