Question

a battery of 9v is connected in series with resistor of 0.2 ohm 0.3 ohm 0.4 Ohm 0.5 and 12 Ohm respectively how much current wood floor throw the 12 ohm resistor​

Answer 1

Answer:

There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as

V= IR

V= IRI= V/R

V= IRI= V/RWhere,

V= IRI= V/RWhere,R is the equivalent resistance of resistances 0.2 Ohm, 0.3 Ohm, 0.4Ohm, 0.5 Ohm and 12 Ohm. These are connected in series. Hence, the sum of the resistances will give the value of R.

R= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ohm

R= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 OhmPotential difference, V= 9 V

R= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 OhmPotential difference, V= 9 VI= 9/13.4 = 0.671 A

R= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 OhmPotential difference, V= 9 VI= 9/13.4 = 0.671 ATherefore, the current that would flow through the 12 Ohm resistor is 0.671 A.

Answer 2

Answer:

0.67 A

EXPLANATION :-

Given :- R 1= 0.2ohm

R2 = 0.3ohm

R3= 0.4ohm

R4=0.5 ohm

R5=12ohm

As it is in series combination so

TOTAL RESISTANCE (R')= R1+R2+R3+R4+R5

R' = (0.2+0.3+0.4+0.5+12)

R'= 13.4

Now V = 9V

FROM OHM'S LAW :-

V = I*R

I = V/R

I = 9/13.4

I = 0.67A

• IN A SERIES COMBINATION OF RESISTORS THE CURRENT REMAINS ...

So , current through 12 ohm resistor = 0.67A