Question

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

Answer 1

Battery=9V

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

Answer:

0.67 A

given that

R1 =0.2 Ohm

R2=0.3 Ohm

R3=0.4 Ohm

R4=0.5 Ohm

R5=12 Ohm

effective resistance in series will be

Rs=r1+r2+r3+r4+r5

 =0.2+0.3+0.4+0.5+12

=13.4

by ohm's law,

V=RI

I=V/R

I=9/13.4

 =0.67 A

Hence current flowing is 0.67 A