a battery of 9V is connected in series with resistors of 0.2ohms ,0.3ohms ,0.4ohms ,0.5ohms and 12 ohms .how much current would flow through the 12 ohms resistors?

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Solution

Total resistance R= 0.2+0.3+0.4+0.5+12= 13.4ohm
potential difference V= 9V
So, current I=V/R = 9 / 13.4
= 0.671 A
Since, it is in series circuit, so current flow will be same
So, current through 12ohm resistor = 0.671 A ampere.

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a battery of 9V is connected in series with resistors of 0.2ohms ,0.3ohms ,0.4ohms ,0.5ohms and 12 ohms .how much current would flow through the 12 ohms resistors?

Total resistance R= 0.2+0.3+0.4+0.5+12= 13.4ohmpotential difference V= 9VSo, current I=V/R = 9 / 13.4= 0.671 ASince, it is in series circuit, so current flow will be sameSo, current through 12ohm resistor = 0.671 A ampere.

Total resistance R= 0.2+0.3+0.4+0.5+12= 13.4ohm
potential difference V= 9V
So, current I=V/R = 9 / 13.4
= 0.671 A
Since, it is in series circuit, so current flow will be same
So, current through 12ohm resistor = 0.671 A ampere.