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A battery of 9V is connected in series with resistors of 0.2 Omega, 0.3 Omega,0.4 Omega, 0.5 Omega and 0.12 Omega respectively. How much current would flow through the 12 Omega resistors?

Answer»

Solution :Here potential difference V=9V
Resistance joined in series are `R_1=0.2 OMEGA, R_2=0.3 Omega, R_3=0.4 Omega, R_4=0.5 Omega and R_5=12 Omega`
`therefore` Total series RESISTANCES `R_s=R_1+R_2+R_3+R_4+R_5`
=0.2+0.3+0.4+0.5+12=`13.4 Omega`
`therefore` Current in the circuit `I=V/R_s= (9V)/(13.4 Omega)=0.67A`
In a series circuit same current flows through all the resistances, hence current of 0.67A will flow through `12 Omega` RESISTOR.


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