Question

1.A bettery of 9 V is connected in series with resistors of 0.2Ω, 0.3Ω, 0.4Ω, 0.5Ω & 12Ω,resp.How much current would flow

through the 12 Ω resistor​

Answer 1

Answer:here is your answer

Explanation: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

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Answer 2

Answer:

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