Series Circuit v.s Parallel Circuit

Series Circuit v.s Parallel Circuit

Series Circuit Parallel Circuit
            
Same current throughout the circuit
i.e I1 = I2 = I3		 The current flowing into and out of the parallel branches is sum of the individual current in the branch.
i.e I1 = I4 = I2 + I3
p.d across the whole circuit is sum of the p.d across each component.
i.e V = V1 + V2		 Same p.d across the parallel branches
i.e V = V1 = V2
The combined resistance for the whole circuit is sum of all the resistances.
i.e Reff = R1 + R2

Note that Reff is always larger than the largest of the individual resistance.		 The reciprocal of the combined resistance is sum of the reciprocal of individual resistance
i.e 1⁄Reff = 1⁄R1 + 1⁄R2 or

Reff = (1⁄R1 + 1⁄R2)-1

Note that Reff is always smaller than the smallest of the individual resistance.
Note that “combined resistance” is also known as “effective resistance” or “equivalent resistance”.

Advantage & Disadvantage of Parallel Circuit

Advantage of Parallel Circuit
  • Light bulbs connected in parallel are brighter than that connected in series.
  • When one of the bulbs in parallel circuit blows, others will continue to glow. However, when one blows in series circuit, the entire circuit will be open. None will glow.

Disadvantage of Parallel Circuit
  • The power source in parallel circuit will deplete out faster than that in series circuit.

Tips in finding Effective Resistance

  • Pay attention to question! Which effective resistance is required! Sometimes, it is not always for the whole circuit. It can be just between two points.
  • Try to draw an equivalent circuit such that all the resistors in straight lines. When redrawing the circuit, simply following through the wire.
Example 1:
Calculate
(a) effective resistance between A and B
(b) I1
(c) p.d across AB
Answer:

(a) If you do not know how to start this, try to redraw the circuit.

Step 1: Start from A, there are two branches after A;
Step 2: One goes to 20 Ω and
      One goes to 25 Ω followed by 35 Ω;
Step 3: These two branches rejoined at B

Hence, that part of circuit is equivalent to
Reff,AB = (1⁄20 + 1⁄(25+35))-1 = 15Ω
(b) The whole circuit is equivalent to 
I1 is the current from power source.
so I1 = V/R = 30/(10+15)) = 1.2 A

(c) p.d across AB = I1 RAB = 1.2 * 15 = 18V

OR

p.d across 10Ω = 10 * 1.2 = 12V
p.d across AB = 30 – 12 = 18V
Example 2:
Find effective resistance
(a) between A and C;
(b) between A and B.
Answer:

(a) The above circuit is equivalent to
Reff,AC = (1⁄(R1+R2) + 1⁄R3 + 1⁄(R4+R5))-1
(b) Main focus is on A and B.

Step 1: Start from A, three branches after A;
Step 2: One goes to R1;
             One goes to R3 and
             One goes to R4 followed by R5.
Step 3: Before these three branches rejoined at B, last two branches rejoined at C first, then followed by R2
Step 4: Branches rejoined at B.

The above circuit is then equivalent to
Reff,AC = (1⁄R3 + 1⁄(R4+R5))-1
Reff,AB = (1⁄R1 + 1⁄(RAC+R2))-1
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