Question

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

Answer 1

Resistance R = 0.2+0.3+0.4+0.5+12
= 13.4
we have V=IR from this
we have I= V/R
= 9/13.4
= 0.67A

Answer 2

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Total resistance of resistors when connected in series is given by

= 1 + 2 + 3 + 4 + 5

= 0.2 Ω+ 0.3 Ω+ 0.4 Ω+ 0.5 Ω+ 12 Ω = 13.4 Ω

According to Ohm’s law,

V = IR ⟹

=/ 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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