Series Circuits

 shows three resistors R₁, R₂ and R₃ connected end to end, i.e., in series, with a battery source of V volts. Since the circuit is closed, a current I will flow and the voltage across each resistor may be determined from the voltmeter readings V₁, V₂ and V₃.

Figure 3.1 : Series circuit

In a series circuit:

(a) the current I is the same in all parts of the circuit; therefore, the same reading is found on each of the two ammeters shown, and,

(b) the sum of the voltages V₁, V₂ and V₃ is equal to the total applied voltage, V, i.e.,

From Ohm’s law:

V₁ =IR₁, V₂ =IR₂, V₃ =IR₃ and V =IR

where R is the total circuit resistance.

Since V =V₁ +V₂ +V₃

then IR =IR₁ +IR₂ +IR₃

Dividing throughout by I gives:

So, for a series circuit, the total resistance is obtained by adding together the values of the separate resistances.

Example 3.1

For the circuit shown in Figure 3.2, determine (a) the battery voltage V, (b) the total resistance of the circuit, and (c) the values of resistance of resistors R₁, R₂ and R₃, given that the voltages across R₁, R₂ and R₃ are 5 V, 2 V and 6 V, respectively.

Figure 3.2 : Circuit for Example 3.1

Solution

(a) Battery voltage V=V₁+V₂+V₃

(b) Total circuit resistance 

(c) Resistance 

Resistance 

Resistance 

(Check: R₁ +R₂ +R₃ = 1.25 + 0.5 + 1.5 = 3.25 Ω =R)

Example 3.2

For the circuit shown in Figure 3.3, determine the voltage across resistor R₃. If the total resistance of the circuit is 100 Ω, determine the current flowing through resistor R₁. Find also the value of resistor R₂.

Figure 3.3 : Circuit for Example 3.2

Solution

Voltage across R₃, V₃ = 25-10-4=11 V

Current  which is the current flowing in each resistor

Resistance 

Example 3.3

A 12 V battery is connected in a circuit having three series-connected resistors having resistances of 4 Ω, 9 Ω and 11 Ω. Determine the current flowing through, and the voltage across the 9 Ω resistor. Find also the power dissipated in the 11 Ω resistor.

Solution

The circuit diagram is shown in Figure 3.4.

Figure 3.4 : Circuit for Example 3.3

Total resistance R=4+9+11=24 Ω

Current  which is the current in the 9 Ω resistor.

Voltage across the 9 Ω resistor, 

Power dissipated in the 11 Ω resistor,