Question

A battery of 9 volt is connected in series with resistors of sur 0.2 ohm, 0.31, 0.4 mm, 0.5 mm and 12 mm respectively how much current will flow through 12 ohm resistor

Answer 1

pd=9v
total resistance in series=0.2+0.31+0.4+0.5+12=_
V=IR {by ohm s law}
I=V/R
calculate on your own

Answer 2

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Total resistance of resistors when connected in series is given by

= r1 + r2 + r3 + r4 + r5

= 0.2 Ω+ 0.31 Ω+ 0.4 Ω+ 0.5 Ω+ 12 Ω = 13.4 Ω

According to Ohm’s law,

V = IR

I =V/R

I = 9 /13.4 = 0.67

▶▶There is no current division occurring in a series circuit.

So, the current through the 12 Ω resistor will be same as 0.67 A.

I hope, this will help you___❤❤

Thank you

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