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

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

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Total resistance  =R

R=0.2+0.3+0.4+0.5+12=13.4\Omega

V = 9V

I=\frac{V}{R}=\frac{9}{13.4}=0.67 A

Hence, All resistors are in series so 0.67A current would flow through the 12\Omega  resistor.                 

Posted by

seema garhwal

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