A network of resistors is connected to a 5 V battery with internal resistance of `1Omega`  as shown in Fig. (a) Compute the equivalent resistance of the network. (6) Obtain the current in each resistor. (c) Obtain the voltage drops `V_(BC)` and `V_(CD)`

Explanation

(a) The network is a simple series and parallel combination of resistors. First two `2Omega` resistors in parallel are equivalent to a resistor   =`(2x2)/(2+2)`=`1Omega`.
In the same way, the `3Omega` and `6Omega` resistors in parallel are equivalent to a resistor of ,
=`(6xx3)/(6+3)`=2Omega`
The equivalent resistance R of the network is obtained by combining these resistors `(2Omega and 4omega)` with `1Omega` in series, that is,
1+1+2+1=`5Omega`
(b) The total current I in the circuit is
I=`(E)/(R)`=`(5)/(5)`=1A
Consider the resistors between A and B. If `I_(1)` is the current in one of the `2Omega` resistors and `I_(2)` the current in the other, `I_(1)xx 2`=`1_(2) xx2` that is, `I_(1)` = `I_(2)` which is otherwise obvious from the symmetry of the two arms. But `I_(1)+1_(2)`=I=1 A. Thus, `1_(1)`= `I_(2)` =`(1)/(2)`  A that is current in each `2Omega` resistor is`(1)/(2) A`. Current in `1Omega` resistor between B and C would be 1 A. Now, consider the resistances between C and D. If `I_(3)` is the current in the `6Omega` resistor, and `I_(4)` in the `3Omega` resistor,`I_(3)xx6`=`I_(4)xx3`i.e, `I_(4)`=`2I_(3)`
But , `I_(3)+I_(4)`=I=1A
Thus, `I_(3)`=`((1)/(3))A`, `I_(4)`=`((2)/(3))A`   
(c ) The voltage drop across AB is `V_(AB)`=`I_(1)xx2=(1)/(2)Axx2Omega`=IV
The voltage drop across BC is `V_(BC)`= `1Axx1Omega`=1V  
Finally , the voltage across CD is `V_(CD)`=`6OmegaxxI_(3)`=`6Omrgaxx((1)/(3))0A`=2V