Find the voltage drop across `18 Omega` resistor in the given circuit.

Explanation

The emf of the source = 30 V
The values of resistances are `18 Omega, 12 Omega, 6 Omega, " and " 6 Omega`
Two resistances `12 Omega` and `6 Omega` are connected in series.
`R_s = 12 + 6 = 18 Omega`
` R_2 = 6 Omega, R_3 = 18 Omega`
The resistances `R_2` and `R_3` are connected in parallel.
The effective resistance is given by
`1/R_p = 1/R_2 + 1/R_3 = 1/6 + 1/18`
`= (3+1)/18 = 4/18 = 2/9`
`:." "R_p= 9/2 Omega`
The resistances `R_1` and `R_p` are connected in series.

`R_1 = 18 Omega, R_p =9/2 Omega`
` R_s = 18 + 9/2 = (36 + 9)/2 = 54/2`
`:." Effective resistance of the circuit " = 45/2 Omega`.
Electromotive force V = 30 V
`:." Current I " = V/R = 30/45/2 = (30xx2)/45`
` = 60//45 = 4/3 amp`.
The voltage drop across `18 Omega` is V = IR
` V = 4/3 xx 18 = 24 V`