Question

A battery of 9v is connected in series with resistors of 0.2 0.3 0.4 0.5 and 12 how much current will flow through

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

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Total resistance of resistors when connected in series is given by

= 1 + 2 + 3 + 4 + 5

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

According to Ohm’s law,

V = IR

I =V/R

I = 9 /13.4 = 0.67

▶▶There is no current division occurring in a series circuit.

So, the current through the 12 Ω resistor will be same as 0.67 A.

I hope, this will help you___❤❤

Thank you

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