Question

A battery of 9v is connected in series with resistors of 0.2 ohm, 0.3ohm, 0.4ohm, 0.5ohm and 12ohm respectively. how much current would flow through 12ohm resistor

Answer 1

Given 
Battery=9V
resistors of 0.2 ohm, 0.3ohm, 0.4ohm, 0.5ohm and 12ohm are connected in series to battery.
R1=0.2ohm
R2=0.3ohm
R3=0.4ohm
R4=0.5ohm
R5=12ohm
As all are connected in series effective resistance would be :
Reff=R1+R2+R3+R4+R5
=0.2+0.3+0.4+0.5+12
=13.4OHMS

By ohms law : V=IR
I=V/R
=9/13.4
=0.67 A

As resistors are connected in series. I=I1=I2=I3=I4=I5
Hence current across 12ohms resistor is 0.67A            

Answer 2

Heya......!!!!

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In this question we have to take out Current that would flow through 12  Ω resistor

=> We know that Current ( I ) = V/R ( By ohm's Law )

V is given to us in question = 9 V .

-- Given in the question :--

• R1 = 0.2 Ω
• R2 = 0.3 Ω
• R3 = 0.4 Ω
• R4 = 0.5 Ω
• R5 = 12 Ω

Battery of 9 v .

As they are connected in series then the effective resistance will be = ( R1 + R2 + R3 + R4 + R5 )

→ 0.2 + 0.3 + 0.4 + 0.5 + 12 => 13.4  Ω

Effective Resistance => 13.4  Ω

and current ( I ) = V/R => 9/13.4

➡ Current ( I ) = 0.67 A

As given in the question that they are connected in series then the current will be same .

So ,, Current that would flow through 12 Ω Resistor is → 0.67 A .

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