Q. 24.8( 21 Votes )
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. How much current would flow through the 12 Ω resistor?
Here a battery of 9 V is connected in series with resistors of R1=0.2ohm, R2=0.4ohm, R3=0.3ohm, R4=0.5ohm, R5=12ohm,
So the resultant resistance = R1 + R2 + R3 + R4 + R5
R = 0.2+0.4+0.3+0.5+12=13.4ohm
As we know that
Thus the current flow through 12ohm resistance will be equal to the current flowing across the whole circuit as in case of Series Connection same current flows through each resistance
∴ I = V/R
I = 9/13.4
I = 0.67amp.
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