Electricity: Combination of Resistors

Numerical Problems

These numerical problems are based on the chapter Electricity class 10 science of NCERT book and CBSE syllabus.

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, R₁ = 10 Ω, R₂ = 20 Ω, R₃ = 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 (R₁) of one electric lamp, M = 15.2 Ω

Resistance (R₂) 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 (V₁) across electric lamp M `= 15.2 Ω ÷ 6.25 A = 2.432 V`

Potential difference (V₂) 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 R₁ and R₂ 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, R₁ = 20 Ω, R₂ = 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