A battery of 9V is connected in series with resistor of 0.2,0.3,0.4,0.5,12 Ohm. How much current would flow through the12 Ohm resistor?

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Solution

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
=13.4 OHMS

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

Thank you :)

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