Series Combination | Resistors are joined from end to end, as if making a queue. |
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Parallel Combination | Resistors are joined in a way that one end of all resistors is joined to positive terminal, and another end of all resistors is joined to negative terminal. |

Resistors are joined in two ways, i.e. in series and in parallel.

When resistors are joined from end to end, it is called the series combination. You can compare such a combination as people standing in a line and one person is holding the hand of the next person and so on. In this case, the total resistance of the system is equal to the sum of the resistance of all the resistors in the system. Bulbs in decorative light for festivals are joined in this fashion.

Following figure shows a circuit in which three resistors R_{1}, R_{2} and R_{3} are joined in series.

Fig: Resistance in Series

Let total resistance = R

Resistance of resistors are R_{1}, R_{2}, R_{3}, … R_{n}

Therefore, R = R_{1} + R_{2} + R_{3} + …………+ R_{n}

When all the resistors are joined in a way that positive ends of all of them are joined together and so do the negative ends, it is called a parallel combination. Electrical appliances in household wiring are joined in htis fashion. When resistors are joined in parallel, the reciprocal of total resistance of the system is equal to the sum of reciprocal of the resistance of resistors. Following figure shows a circuit in which three resistors R_{1}, R_{2} and R_{3} are joined in parallel.

Fig: Resistance in Parallel

Let total resistance = R

Resistance of resistors are R_{1}, R_{2}, R_{3}, … R_{n}

`1/R=(1)/(R_1)+(1)/(R_2)+(1)/(R_3)+…….+(1)/(R_n)`

Example 1: There are three resistors joined in series in a system having resistance equal to 10 Ω, 20 Ω and 30 Ω respectively. If the potential difference of the circuit is 240 V, find the total resistance and current through the circuit.

**Solution:** Given, R_{1} = 10 Ω, R_{2} = 20 Ω, R_{3} = 30 Ω and V = 240 V

Total resistance (R) =?

Current through the circuit ( I ) =?

According to Ohm's Law Total resistance in series (R) = Sum of resistance of all resistors

Or, `R = 10 Ω + 20 Ω + 30 Ω = 60 Ω`

We know that electric current `I = V/R`

Or, `I = 240 V ÷ 60 Ω = 4 A`

Thus, total resistance `(R)= 60 Ω`

Current through the circuit = 4 A

Example 2: There are two electric lamps M and N which are joined in a series having resistance equal to 15 Ω and 20 Ω respectively. If the potential difference between two terminals of electric circuit is 220V, find the total resistance and electric current through the circuit. Also find the potential difference across the two lamps separately.

**Solution:** Given, resistance (R_{1}) of one electric lamp, M = 15.2 Ω

Resistance (R_{2}) of other electric lamp, N = 20 Ω

Potential difference (V) through the circuit = 220 V

Electric current (I) through the circuit =?

Potential difference through each of the electric lamp =?

According to Ohm’s Law; total resistance in series

= Sum of resistance of all resistors `= 15.2 Ω + 20 Ω = 35.2 Ω`

Electric Current `I = V/R = 220 V ÷ 35.2 Ω = 6.25 A`

Potential difference (V_{1}) across electric lamp M `= 15.2 Ω ÷ 6.25 A = 2.432 V`

Potential difference (V_{2}) across electric lamp N `= 20 Ω ÷ 6.25 A = 3.2 V`

Thus, electric current through the circuit = 6.25 A

Potential difference through electric lamp M = 2.432 V

Potential difference through electric lamp N = 3.2 V

Example 3: There are two resistors R_{1} and R_{2} having resistance equal to 20Ω and 30Ω respectively are connected in parallel in an electric circuit. If the potential difference across the electric circuit is 5 V, find the electric current flowing through the circuit and the total resistance of the resistors.

**Solution:** Given, R_{1} = 20 Ω, R_{2} = 30 Ω, Potential difference (V) = 5 V

Total resistance (R) =?

Electric current (I) through the circuit =?

We know that in parallel combination, the reciprocal of total resistance is,

`1/R=(1)/(R_1)+(1)/(R_2)`

`=(1)/(20Ω)+(1)/(30Ω)`

`=(3+2)/(60)Ω=(5)/(60)Ω`

Or, `1/R=(1)/(12)Ω`

Now, electric current through the circuit `I = V/R`

Or, `I = 5 V ÷ 12 Ω = 0.416 A`

Thus, total resistance `R = 12 Ω`

Electric current (I) through the circuit = 0.416 A

Example 4: There are five electric appliances, viz. electric heater and electric lamp, an electric fan, computer and an exhaust fan are connected in parallel in a household. The resistance electric appliances are 40Ω, 5 Ω, 8Ω, 20Ω and 10Ω respectively. If an electric current of 240V is flowing through the circuit then find

- Total resistance through the circuit
- Total electric current (I) through the circuit and
- the current through each of the resistor.

**Solution:**We know that the recpirocal of total resistance = sum of reciprocals of individual resistances

`(1)/(R_text(total))=(1)/(40Ω)+(1)/(5Ω)+(1)/(8Ω)+(1)/(20Ω)+(1)/(10Ω)`

`=(1+8+5+2+4)/(40)Ω`

`=(20)/(40)Ω=1/2Ω`

Or, `R_text(total)=2 Ω`

Now, electric current through the circuit `= V/R`

Or, `I = 240 V ÷ 2 Ω = 120 A`

Similarly, electric current through different appliances can be calculated as follows:

Electric current through heater `= 240 V ÷ 40 Ω = 6 A`

Electric current through electric lamp `= 240 V ÷ 5 Ω = 48 A`

Electric current through electric fan `= 240 V ÷ 8 Ω = 30 A`

Electric current through computer `= 240 V ÷ 20 Ω = 12 A`

Electric current through exhaust `= 240 V ÷ 10 Ω + 24 A`

The total current through the circuit can be calculated by adding the electric current through individual resistors.

Or, I_{total} = 6 A + 48 A + 30 A + 12 A + 24 A = 120 A

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