Question

a battery of 9 volt is connected in series with resistors of 0.0 and 0.3 and 0.4 and 0.518 M and 12 respectively how much current will flow through the 12th and 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

I= V/R

Where,

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

Potential difference,

V= 9 V

I= 9/13.4

= 0.671 A

Therefore, the current that would flow through the 12 Ohm resistor is 0.671 A.

Explanation:

Hope this helps you

Answer 2

Answer:

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            

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Explanation: