Poker battery of 9 volt is connected in parallel with resistors of 0.2 ohm and 12 ohm how much current will flow through that 12 ohm resistor
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hi
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
hope it helps(:
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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tumpachoudhuri2057:
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