Of course! Let’s suppose that we are asked to solve the following problems.
The circuit consists of three resistors of 30, 40, and 50 Ω connected in series. What is the resistance of the circuit?
Common resistance for series connection:
R=R1+R2+R3=30+40+50=120 Ω.
The circuit consists of five resistors r=30 kΩ connected in series. What is the total resistance?
R=n·r=5·30·〖10〗^3=150·〖10〗^3 Ω.
The circuit has three resistors of 100, 300 and 35 Ω connected in parallel. What is the total resistance?
For parallel connection:
1/R=1/R1+1/R2+1/R3 ⇨ R=(R1·R2·R3)/(R1R2+R2R3+R1R3)=
=(100·300·35)/(100·300+300·35+35·100)=23.9 Ω.
The circuit consists of 5 equal resistors of r=2 kΩ each. What is the total resistance?
According to the previous formula for equal resistors:
R=r/n=(2·〖10〗^3)/5=400 Ω.
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