In the circuit `R_(1) = 4 Omega , R_(2) = R_(3) = 15 Omega R_(4) = 30 Omega and E = 10 V` Calculate the equivalent resistance of the circuit and current in each resistor

Answer

(`6)/(30) A`

Explanation

`R_(2) ,R_(3) and R_(4)` are in parallel their effective resistance `R_(p)` is given by

`(1)/(R_(p)) = (1)/(R_(2)) + (1)/(R_(3)) + (1)/(R_(4)) = (1)/(15)+ (1)/(15)+ (1)/(30)+ (5)/(30)+ (1)/(6)`
or `R_(P) = 6 Omega`
Now equivalent resistance of the circuit is
`R = R_(1) + R_(p) = 4 + 6 = 10 Omega`
current `I_(1) = (10)/(10) = 1A`
pot diff across `A and B = I_(1) R_(p) = 1 xx 6 = 6 V`
Current through `R_(2) or R_(3), I_(2) = (6)/(15) A , l_(3) =(6)/(15) A`
Current through `R_(4), I_(4)= (6)/(30) A`