Resistors in SERIES configuration

When resistors are connected end-to-end so that there is only one path for the flow of electric current, the resistors are said to be series connected.

The effective resistance of a series circuit depends on the number and the value of resistors connected in the circuit. The more resistors connected in series, the higher the effective resistance of the circuit. The higher the resistance of the circuit the more the opposition there is to the flow of electric current.

Below are the rules to be followed in computing the unknown quantities in series circuit.

1. The total resistance (RT) is equal to the sum of the individual resistances.
   R_T = R₁ + R₂ + R₃ + ... + R_N
   
   
2. The current is the same throughout the circuit.
   I_T = I₁ = I₂ = I₃ = ... = I_N
   
   
3. The total voltage (ET) is equal to the sum of individual voltage drops across each resistor.
   E_T = E₁ + E₂ + E₃ + ... + E_N
   
   
4. The total power (PT) is equal to the sum of individual power dissipated by each resistor.
   P_T = P₁ + P₂ + P₃ + ... + P_N

The numerical subscripts refer to the individual resistance, current, voltage or power of the circuit. The subscript “T” represents the total or effective resistance, current, voltage or power of the circuit.


EXAMPLE 1:

Resistors in series circuit

GIVEN:
ET = 100
R1 = 50 ohms
R2 = 150 ohms

REQUIRED: RT IT E1 E2

PT P1 P2

SOLUTION:
Compute for RT.		RT = R1+R2 RT = 50Ω+150Ω RT = 200Ω
After getting RT, calculate the total current using ohm’s law.		IT = ET/RT IT = 100v/200Ω IT = 0.5A
Since the individual current and resistance are given, you can now solve for the value of individual voltage using ohm’s law.		E1 = I1R1 E1 = (0.5A)(50Ω) E1 = 25V E2 = I2R2 E2 = (0.5A)(150Ω) E2 = 75V
Proving the rule for voltage.		ET = E1 + E2 100V = 25V + 75V 100V = 100V
Solve for the value of PT and individual powers P1 and P2 using the power formula P = IE		PT = IT ET PT = (0.5A)(100V) PT = 50W P1 = I1E1 P1 = (0.5A)(25V) P1 = 12.5W P2 = I2 E2 P2 = (0.5A)(75V) P2 = 37.5W
Proving the rule for power.		PT = P1 + P2 50W = 12.5W + 37.5W 50W = 50W

----------------------------------------------------------------------------------------

EXAMPLE 2:

Resistors in series configuration

GIVEN:

IT = 2A
R1 = 25Ω
R2 = 50Ω
R3 = 75Ω 

REQUIRED: RT ET E1 E2 E3

PT P1 P2 P3

SOLUTION:
Compute for the value of RT		RT = R1+R2+R3 RT = 25Ω+50Ω+75Ω RT = 150Ω
Solve for the total voltage (ET) of the circuit using ohm’s law.		ET = (IT)(RT) ET = (2A)(150Ω) ET = 300V
Current is the same throughout the circuit.		IT = I1 = I2 = I3 = 2A
Solve for the individual voltage using ohm’s law.		E1 = I1R1 E1 = (2A)(25Ω) E1 = 50V E2 = I2R2 E2 = (2A)(50Ω) E2 = 100V E3 = I3R3 E3 = (2A)(75Ω) E3 = 150V
Proving the rule for voltage in series circuit.		ET = E1 + E2 +E3 300V = 50V + 100V + 150V 300V = 300V
Compute for PT and individuals power, P1, P2 and P3.		PT = IT ET PT = (2A)(75A) PT = 600W P1 = I1E1 P1 = (2A)(50V) P1 = 100W P2 = I2E2 P2 = (2A)(100V) P2 = 200W P3 = I3E3 P3 = (2A)(150V) P3 = 300W
Proving the rule for power in series circuit.		PT = P1 + P2 +P3 600W = 100W + 200W + 300W 600W = 600W