In the circuit diagram given below five resistances of 10 Ω, 40 Ω, 30 Ω, 20 Ω, and 60 Ω are connected as shown to a 12 V battery.

Calculate:

(*a*) total resistance in the circuit.

(*b*) total current flowing in the circuit.

#### Solution

(a) When three resistors are connected in parallel, the net resistance can be obtained as followed:

1/*R* = 1/*R*_{1} + 1/*R*_{2} + 1/*R*_{3}

The resistors of 30 Ω, 20 Ω and 60 Ω are connected in parallel. Therefore, the net resistance *R* will be:

1/*R* = 1/30 + 1/20 +1/60

1/*R* = (2 + 3 + 1)/60

1/*R* = 6/60*R* =10 Ω

Two more resistors of 10 Ω and 40 Ω are connected in parallel to each other. Therefore, their net resistance, *R'* will be:

1/*R ^{'}* = (1/10) + (1/40)

1/

*R*= (4 + 1)/40

^{'}1/

*R*= 5/40

^{'}*R'*= 8 Ω

The resistors of 8 Ω and 10 Ω are connected in series. Therefore, the net resistance of the circuit =R+R

^{,}=8 Ω + 10 Ω = 18 Ω

(b) The total current flowing through the circuit can be calculated as:

*I*=

*V*/

*R*

*I*= 12/18

*I*= 0.67 A