R1=6 Ω
R2=3 Ω
R3=2 Ω
R4=4 Ω
6 Ω and 3Ω resistances are connected in parallel.
First, find the equivalent resistance:
ER1 = R1*R2/(R1+R2)= 6*3/6+3 = 18/9 = 2Ω
The resistances of 2 Ω and 4 Ω are connected in series:
TR2 = R3+R4 = 2 Ω+4 Ω= 6 Ω (b)
ER1 and TR2 are connected in parallel.
RT = ER1*TR2/ ER1+TR2= 2*6/2+6=12/8 = 1.5 Ω
Now, we can calculate the I (current)
On parallel circuits, Voltage is always the same. (12V)
But for series is VT=V1+V2
We have to find the current flowing across R3 and R4 (TR2)
V= I*R
I= V/R = 12/TR2= 12/6 = 2 A
With that current, we can calculate the voltage (potential difference) on R3 (2 Ω)
V=I*R= 2*2 = 4V (ai)
(aii) potential difference across the 3 Ω resistor
Since it is a parallel circuit , potential difference is always the same = 12V
b.combined resistance of the 2 Ω and 4 Ω resistors connected in series
R3+R4 = 2 Ω+4 Ω= 6 Ω (b)