A conductor of resistance `20 Omega` having uniform cross sectional area is bent in the form of a closed ring. A cell of emf 1.5 V and of negligible internal resistance is joined to the ring between two points dividing the circumference of the ring in the ratio 3:1 Find the currents flowing through the two parts of the ring .

Explanation

Resistance of the part ACB=`20 times 3/4 =15 Omega`
Resistance of the part ADB=`20 times 1/4 =5 Omega`
`therefore` Equivalent resistance of the circuit,
`R=(15times5)/(15+5)=75/20=3.75 Omega`

emf of the cell ( E)=1.5 V
`therefore` Main current in the circuit ( l)=`1.5/3.75=0.4 A`
`therefore` Current through ACB (`I_1`)=`5/20 times 0.4 =0.1 A`
Current through ADB (`I_2`)=`15/20times0.4=0.3A`