A battery of 9 V is connected in series with resistors of 0.2 Ohms, 0.3 Ohms, 0.4 Ohms, 0.5Ohms and 12Ohms respectively. How much current would flow through the 12 Ω resistors?

 Given

A battery of 9 V is connected in series with resistors of 0.2 Ohms, 0.3 Ohms, 0.4 Ohms, 0.5Ohms and 12Ohms respectively

Find out

Amount of the current flow through the 12 Ω resistors

Solution

According to Ohm’s law

V= IR

Therefore, 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=  R1 + R2 + R3 + R4 + R5

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

The potential difference, V= 9 V

I = 9/13.4 = 0.671 A

When resistors are connected in series, the current is the same in all the resistors. Hence, current in 12 Ω resistor 0.67 A

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