Question

a battery of 9 volt is connected in series with resistors of 0.2 M 0.3 mm 0.4 ohm 0.5 ohm and 12 ohm respectively how much current will flow through the 12 ohm 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

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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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