50 Numerical Questions On Electricity Class 10

50 Numerical Questions

On Electricity Class 10 with Solutions

Source: omtexclasses.com

Numerical Problems based on Electric Current

This provides a comprehensive guide on solving numerical problems related to electricity, including electric current, potential difference, Ohm's law, resistance, and resistivity. It presents various formulas and examples to calculate charge, current, voltage, and resistance in different scenarios. Additionally, it covers the combination of resistances in series and parallel configurations.

Formulas to Remember

To solve the numericals on electricity, we will use the following formulas:

$$ I = \frac{Q}{t} $$

$$ I = \frac{ne}{t} $$

Where:
I = Electric current – Ampere
Q = charge – Coulomb
T = time - second
N = number of electrons
E = charge on an electron = \( 1.6 \times 10^{-19} \)

1. The filament of bulb draws a current of 0.5 ampere. Calculate the amount of charge if bulb glows for 2.5 hrs.

Solution

Given values Electric current (I) = 0.5 A
Time (t) = 2.5 Hrs = \( 2.5 \times 60 \times 60 = 9000 \) sec
Charge (Q) = ?

$$ I = \frac{Q}{t} $$

$$ Q = It $$

$$ Q = 0.5 \times 9000 $$

Q = 4500 Coulomb

2. 10 Coulombs charge passing through a point in a circuit in 5 seconds. Find the electric current flowing in the circuit.

Solution

Given values Time (t) = 5 sec
Charge (Q) = 10 C
Electric current (I) = ?

$$ I = \frac{Q}{t} $$

$$ I = \frac{10}{5} $$

I = 2 Amp.

3. A current of 0.5 Ampere is drawn by a filament of an electric bulb for 15 minutes. Find the amount of electric charge.

Solution

Given values Electric current (I) = 0.5 A
Time (t) = 15 min = \( 15 \times 60 = 900 \) sec
Charge (Q) = ?

$$ I = \frac{Q}{t} $$

$$ Q = I \times t $$

$$ Q = 0.5 \times 900 $$

Q = 450 Coulomb

4. A current of 1 Ampere is drawn by a filament of an electric bulb. Find the Number of electrons passing through a cross section of the filament in 10 seconds.

Solution

Given values Electric current (I) = 1 A
Time (t) = 15 s
Number of electrons (n) = ?
Charge on electrons (e) = \( 1.6 \times 10^{-19} \) C

$$ I = \frac{ne}{t} $$

After cross multiplication, we find:

$$ n = \frac{I \times t}{e} $$

Now put the given values in the formula:

$$ n = \frac{1 \times 10}{1.6 \times 10^{-19}} $$

$$ n = \frac{100}{16 \times 10^{-19}} $$

$$ n = \frac{6.25}{10^{-19}} $$

\( n = 6.25 \times 10^{19} \)

Number of electrons (n) has no unit required.

5. 150000 coulomb charge is required to deposit one mole of copper from the CuSO4 solution. Find the time to deposit 0.2 mole copper from the copper solution if a current of 5 Ampere is flowing the solution.

Solution

Given values Electric current (I) = 5 Amp
Charge (Q) = 150000 C

Charge required for 1 mole copper = 150000 C

Charge required for 0.2 mole copper = \( 150000 \times 0.2 = 30000 \) C

$$ I = \frac{Q}{t} $$

$$ t = \frac{Q}{I} = \frac{30000}{5} $$

t = 6000 S

6. An heater draws a current 10 A for 5 minutes. Calculate the charge flowing in heater.

Solutions

Given values Electric current (I) = 10 A
Time (t) = 5 min = \( 5 \times 60 = 300 \) s
Charge (Q) = ?

$$ I = \frac{Q}{t} $$

$$ Q = I \times t $$

$$ Q = 10 \times 300 $$

Q = 3000 C

7. \( 1.25 \times 10^{18} \) electrons are passed from one end to another end of a conductor in 10 seconds. Find the current flowing through the conductor.

Solution

Given values Number of electrons (n) = \( 1.25 \times 10^{18} \)
Time (t) = 10 s
Electric current (I) = ?

$$ I = \frac{ne}{t} $$

[e = \( 1.6 \times 10^{-19} \)]

$$ I = \frac{1.25 \times 10^{18} \times 1.6 \times 10^{-19}}{10} $$

I = 0.2 A

8. Calculate the number of electrons present in 0.1 coulomb of charge.

Solution

Given values Charge (Q) = 0.1
Number of electrons (n) = ?

Number of elections in 1 Coulombs is = \( 6.25 \times 10^{18} \)

Number of electrons in 0.1 Coulomb charge is = \( 6.25 \times 10^{18} \times 0.1 \)

= \( 6.25 \times 10^{17} \)

Numerical Problems based on Potential difference and work

Formulas to Remember

$$ V = \frac{W}{Q} $$

Where:
V = Potential difference – volt
W = work done - Joule
Q = charge – Coulomb

9. Calculate the amount of work done in carrying 8C charge from a terminal of 150 volt to 200 volt.

Solution

Given values Charge (Q) = 8 C
Potential (V1) = 150 volt
Potential (V2) = 200 volt

Potential difference (V) = V2 - V1

200 – 150 = 50 volt

Work (W) = ?

$$ V = \frac{W}{Q} $$

$$ W = VQ $$

$$ W = 50 \times 8 = 400 J $$

10. How much energy is given to each coulomb of charge passing through a 10 Volt battery?

Solution

Given values Potential difference/voltage (V) = 10 volt
Charge (Q) = 1 C
Energy (E) = ?

$$ V = \frac{W}{Q} $$

Work is defined as energy transferred by force, so We can put E in the place of W

$$ E = VQ $$

= 10 × 1 = 10 J

11. Work done in moving 5 coulomb charge from point A to B is 80 Joule, if potential at point A is 25 volt then find the potential at B.

Solution

Given values Work (W) = 80 J
Charge (Q) = 5 C
Potential (VA) = 25 volt
Potential (VB) = ?

$$ V = V_B - V_A = \frac{W}{Q} $$

$$ V_B - 25 = \frac{80}{5} $$

$$ V_B - 25 = 16 $$

$$ V_B = 16 + 25 = 41 \text{ volt} $$

Numerical Problems based on Ohm’s law

Formulas to Remember

$$ V = IR $$

$$ R = \frac{V}{I} $$

$$ I = \frac{V}{R} $$

Where:
V = Voltage (Potential difference) = volt
R = Resistance – Ohm (Ω)
I = Electric current – Ampere

12. When a cell of 1.5 volt is applied in a circuit, a current of 0.5 ampere flows through it. Calculate the resistance of the circuit.

Solution

Given values Voltage (V) = 1.5 volt
Current (I) = 0.5 A
Resistance (R) = ?

$$ R = \frac{V}{I} $$

Put the values in the formula of resistance:

$$ R = \frac{1.5}{0.5} = 3 \text{ Ohm} $$

13. How much current will an electric bulb draw from a 220-volt source. If the resistance of the filament is 1200 ohm?

Solution

Given values Voltage (V) = 220 volt
Resistance (R) = 1200 Ohm
Current (I) = ?

$$ I = \frac{V}{R} $$

$$ I = \frac{220}{1200} = 0.183 A $$

14. What is the potential difference between the ends of a conductor of 15 Ohm resistance when a current of 2 ampere flows through it.

Solution

Given values Resistance (R) = 15 Ohm
Electric current (I) = 2 A
Potential difference (V) = ?

$$ V = IR $$

$$ V = 2 \times 15 = 30 \text{ volt} $$

15. A heater draws a current of 5 A when it is connected to 110 volts. What current will the heater draw when it is connected to 220 volts.

Solution

Given values Current (I1) = 5 A
Voltage (V1) = 110 volts
Voltage (V2) = 220 volts
Current (I2) = ?

According to Ohm’s law, V=IR

$$ R = \frac{110}{5} = 22 \text{ Ohm} $$

When the voltage is 220 volt, The current will be:

$$ I_2 = \frac{220}{22} = 10 A $$

16. Calculate the potential difference required across a conductor of resistance 6 ohm to make a current of 2.5 A flow through it.

Solution

Given values Resistance (R) = 6 ohm
Current (I) = 2.5 A
Potential difference (V) = ?

$$ V = IR $$

$$ V = 2.5 \times 6 = 15 \text{ volt} $$

Numerical Problems based on Resistance and Resistivity

Formulas to Remember

$$ R = \rho \frac{l}{A} $$

Where:
R = Resistance - Ohm
\( \rho \) = Resisitivity - Ohm m
l = length – m
A = area of cross section - m²

17. Calculate the resistance of a copper wire of length 30 cm and area of cross section \( 3 \times 10^{-4} \text{ m}^2 \). The resistivity of copper is \( 1.7 \times 10^{-8} \text{ ohm m} \).

Solution

Given values Length of wire (l) = 30 cm = 0.3m
Area of cross section (A) = \( 3 \times 10^{-4} \text{ m}^2 \)
Resistivity of copper (\( \rho \)) = \( 1.7 \times 10^{-8} \text{ ohm m} \)
Resistance (R) = ?

$$ R = \rho \frac{l}{A} $$

$$ R = 1.7 \times 10^{-8} \times \frac{0.3}{3 \times 10^{-4}} $$

$$ R = \frac{0.51 \times 10^{-8}}{3 \times 10^{-4}} $$

$$ R = \frac{0.17 \times 10^{-8}}{10^{-4}} $$

$$ R = 0.17 \times 10^{-8} \times 10^4 $$

R = \( 0.17 \times 10^{-4} \text{ Ohm} \)

18. Calculate the resistivity of wire having length 1 m and area of cross section \( 1.20 \times 10^{-6} \text{ m}^2 \), if its resistance is 0.013 ohm.

Solution

Given values Area of cross section (A) = \( 1.20 \times 10^{-6} \text{ m}^2 \)
Length (l) = 1m
Resistance (R) = 0.013 ohm
Resistivity (\( \rho \)) = ?

$$ R = \rho \frac{l}{A} $$

Now put the values in the formula:

$$ 0.013 = \rho \frac{1}{1.20 \times 10^{-6}} $$

$$ \rho = 0.013 \times \frac{1.20 \times 10^{-6}}{1} $$

\( \rho = 1.56 \times 10^{-8} \text{ ohm m} \)

19. Calculate the area of cross section of a wire of 1 m and resistance 25 ohm, if the resistivity of material of the wire is \( 1.84 \times 10^{-6} \text{ ohm m} \).

Solution

Given values Length (l) = 1m
Resistance (R) = 25 ohm
Resistivity (\( \rho \)) = \( 1.84 \times 10^{-6} \text{ ohm m} \)
Area of cross section (A) = ?

$$ R = \rho \frac{l}{A} $$

$$ 25 = 1.84 \times 10^{-6} \times \frac{1}{A} $$

$$ A = 1.84 \times 10^{-6} \times \frac{1}{25} $$

A = \( 7.36 \times 10^{-8} \text{ m}^2 \)

20. 1 m long wire with resistance 0.85 ohm and diameter 0.2 mm, what will be the resistivity of the metal at 20°C.

Solution

Given values Length (l) = 1m
Resistance (R) = 0.85 ohm
Diameter (d) = 0.2 mm = \( 2 \times 10^{-4} \text{m} \)
Radius (r) = d/2 = \( 1 \times 10^{-4} \text{m} \)
Area of cross section (A) = ? (\( \pi r^2 \))

$$ R = \rho \frac{l}{A} $$

$$ R = \rho \frac{l}{\pi r^2} $$

$$ 0.85 = \rho \frac{7}{22 \times (1 \times 10^{-4})^2} \times 1 $$

$$ \rho = \frac{0.85 \times 22}{7 \times 10^{-8}} $$

\( \rho = 2.67 \times 10^{-8} \text{ ohm m} \)

Numerical Problems based on the Combination of Resistances

Formulas to Remember

$$ R = R_1 + R_2 + R_3 + \dots R_n $$

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots \frac{1}{R_n} $$

Where:
\( R_1, R_2, R_3 \dots R_n \) are different resistances
R = total resistance of the combination

21. Three resistances of 5ohm, 10 ohm and 15 ohm are connected in series, find the total resistance.

Solution

Given value R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
R = ?

$$ R = R_1 + R_2 + R_3 $$

$$ R = 5 + 10 + 15 = 30 \text{ ohm} $$

22. Three resistances of 5ohm, 10 ohm and 15 ohm are connected in parallel, find the total resistance.

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
R = ?

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$

$$ \frac{1}{R} = \frac{1}{5} + \frac{1}{10} + \frac{1}{15} $$

$$ \frac{1}{R} = \frac{6+3+2}{30} = \frac{11}{30} $$

$$ \frac{1}{R} = \frac{11}{30} $$

$$ R = \frac{30}{11} = 2.72 \text{ ohm} $$

23. Three resistances of 5ohm, 10 ohm and 15 ohm are connected in series, and system is connected with 90 volt battery. What will the current flowing in the circuit.

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
V = 90 volt
I = ?

$$ I = \frac{V}{R} $$

We don’t have the value of ‘R’ so first we will find the value:

$$ R = 5 + 10 + 15 = 30 \text{ ohm} $$

$$ I = \frac{V}{R} = \frac{90}{30} = 3 A $$

24. Three resistances of 5ohm, 10 ohm and 15 ohm are connected in parallel combination and the circuit is connected with 100 volt battery. Find electric current flowing in each of the resistance.

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
V = 100 volt
I1 = ?
I2 = ?
I3 = ?

$$ I_1 = \frac{V}{R_1} $$

$$ I_1 = \frac{100}{5} = 20 A $$

$$ I_2 = \frac{V}{R_2} $$

$$ I_2 = \frac{100}{10} = 10 A $$

$$ I_3 = \frac{V}{R_3} $$

$$ I_3 = \frac{100}{15} = 6.66 A $$

(Note: Current in all resistors is different in the parallel combination.)

25. Three resistors of 5ohm, 10 ohm and 15 ohm are connected in series combination and 10 ampere current is flowing in the system when connected with 110 Volt. Find the potential difference between two ends of each resistor.

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 15 ohm
V = 110 volt
I = 10 A
V1 = ?
V2 = ?
V3 = ?

$$ V_1 = I R_1 $$

$$ V_1 = 10 \times 5 = 50 \text{ volt} $$

$$ V_2 = I R_2 $$

$$ V_2 = 10 \times 10 = 100 \text{ volt} $$

$$ V_3 = I R_3 $$

$$ V_3 = 10 \times 15 = 150 \text{ volt} $$

26. Three resistances of 2 ohm, 3 ohm and 6 ohm are connected in series and then in parallel. Find the total resistance in both the arrangements.

Solution

Given values R1 = 2 ohm
R2 = 3 ohm
R3 = 6 ohm
R = Total resistance = ?

For series combination:

$$ R = R_1 + R_2 + R_3 $$

$$ R = 2 + 3 + 6 = 11 \text{ ohm} $$

For parallel combination:

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$

$$ \frac{1}{R} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} $$

$$ \frac{1}{R} = \frac{6+4+2}{12} = \frac{12}{12} $$

R = 1 ohm

27. A 18 Volt battery is connected across a lamp whose resistance is 50 ohm, through a variable resistor. If the current flowing through the circuit is 0.3 A. Calculate the value of resistance used from the variable resistor.

Solution

Given values Voltage (V) = 18 volt
Current (I) = 0.3 A
Resistance of bulb (r) = 50 ohm
R1 = resistance form variable resistor = ?

Let R1 is the resistance from variable resistor that is used.

Both R1 and r are is series combination so the resultant resistance is:

$$ R = R_1 + r $$

According to Ohm’s law V=IR

$$ 15 = 0.3 \times (R_1 + r) $$

$$ R_1 + 50 = \frac{18}{0.3} = 60 $$

$$ R_1 = 60 - 50 = 10 \text{ ohm} $$

Numerical Problems based on the Circuit diagrams

28. Calculate the total resistance of the following circuit.

[Insert Image from Page 25 here: Series circuit with 5Ω, 10Ω, 20Ω]

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 20 ohm
R = ?

All resistances in the given circuit are connected in the series so total resistance:

$$ R = R_1 + R_2 + R_3 $$

$$ R = 5 + 10 + 20 = 35 \text{ ohm} $$

29. Calculate the total resistance in the given circuit

[Insert Image from Page 26 here: Parallel circuit with 5Ω, 10Ω, 20Ω]

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 20 ohm
R = ?

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$

$$ \frac{1}{R} = \frac{1}{5} + \frac{1}{10} + \frac{1}{20} $$

$$ \frac{1}{R} = \frac{4+2+1}{20} = \frac{7}{20} $$

$$ R = \frac{20}{7} = 2.85 \text{ ohm} $$

30. You have been given the following circuit diagram. Find the following:
(i) Electric current through each resistor
(ii) Total resistance
(iii) Total current

[Insert Image from Page 27 here: Parallel circuit with 5Ω, 10Ω, 30Ω connected to 6V]

Solution

Given values R1 = 5 ohm
R2 = 10 ohm
R3 = 30 ohm
V = 6 volt
R = ?
I1 = ?, I2 = ? and I3 = ?
I = ?

All the resistors are connected in parallel combination.

(i) Current through each resistor

$$ V = I_1 R_1 $$

$$ 6 = I_1 \times 5 $$

$$ I_1 = \frac{6}{5} = 1.2 A $$

$$ V = I_2 R_2 $$

$$ 6 = I_2 \times 10 $$

$$ I_2 = \frac{6}{10} = 0.6 A $$

$$ V = I_3 R_3 $$

$$ 6 = I_3 \times 30 $$

$$ I_3 = \frac{6}{30} = 0.2 A $$

(ii) Total resistance (R)

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$

$$ \frac{1}{R} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30} $$

$$ \frac{1}{R} = \frac{6 + 3 + 1}{30} = \frac{10}{30} $$

$$ R = \frac{30}{10} = 3 \text{ Ohm} $$

(iii) Total current (I)

$$ V = IR $$

$$ 6 = I \times 3 $$

$$ I = \frac{6}{3} = 2 A $$

31. Find the equivalent resistance in the given circuit and current flowing through the circuit.

[Insert Image from Page 29 here: 3Ω and 6Ω in parallel, connected to 4.5V]

Solution

Given values R1 = 3 ohm
R2 = 6 ohm
V = 4.5 volt
R = ?

We can see that the resistors of 3 ohm and 6 ohm are connected in parallel combination so the equivalent resistance R

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} $$

$$ \frac{1}{R} = \frac{1}{3} + \frac{1}{6} $$

$$ \frac{1}{R} = \frac{2 + 1}{6} = \frac{3}{6} $$

$$ R = \frac{6}{3} = 2 \text{ ohm} $$

According to Ohm’s law, V=IR

$$ 4.5 = I \times 2 $$

$$ I = \frac{4.5}{2} = 2.25 A $$

32. You have been given the following circuit of resistors connected with a battery.
Calculate:
(i) total resistance of the circuit
(ii) total current
(iii) voltage across 5 ohm resistor

[Insert Image from Page 31 here: 10Ω and 10Ω in parallel, connected in series with 5Ω, V=6V]

Solution

Given values R1 = 10 ohm
R2 = 10 ohm
R3 = 5 ohm
V = 6 volt

(i) total resistance

We can see R1 and R2 are in parallel so the total resistance RA

$$ \frac{1}{R_A} = \frac{1}{R_1} + \frac{1}{R_2} $$

$$ \frac{1}{R_A} = \frac{1}{10} + \frac{1}{10} $$

$$ \frac{1}{R_A} = \frac{1+1}{10} = \frac{2}{10} $$

$$ R_A = \frac{10}{2} = 5 \text{ Ohm} $$

Now RA and R3 in series so the total resistance R

R = RA + R3

R = 5 + 5 = 10 ohm

(ii) total current

$$ I = \frac{V}{R} = \frac{6}{10} = 0.6 A $$

(iii) Voltage across 5 ohm resistor

V1 = IR1

V1 = 0.6 × 5 = 3 volt

33. Find the equivalent resistance of the following circuit of resistors.

[Insert Image from Page 33 here: Triangle configuration. 4Ω on one side, 8Ω and 8Ω on others]

Solution

Given values R1 = 4 Ohm
R2 = 8 Ohm
R3 = 8 Ohm
R = ?

In the given circuit diagram we can see R2 and R3 are series so:

RA = R2 + R3

RA = 8 + 8 = 16 Ohm

Now the RA is parallel to R1 so the equivalent resistance R

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_A} $$

$$ \frac{1}{R} = \frac{1}{4} + \frac{1}{16} $$

$$ \frac{1}{R} = \frac{4+1}{16} = \frac{5}{16} $$

$$ R = \frac{16}{5} = 3.2 \text{ Ohm} $$

Numerical Problems based on Work, Power and Energy

Formulas to Remember

$$ V = \frac{W}{Q} $$

$$ W = V \times Q $$

$$ W = I^2Rt $$

$$ W = \frac{V^2t}{R} $$

Note = We can put E in place of W

$$ P = \frac{W}{t} $$

$$ P = \frac{V^2}{R} $$

$$ P = I^2R $$

Where:
P = power – Watt
W = work – Joule
V = voltage – Volt
I = current – Ampere
Q = charge = Coulomb

34. An electric bulb is connected to 110 volt electric source. The current is 0.5 A, then what is the power of the bulb?

Solution

Given values Voltage (V) = 110 volt
Current (I) = 0.5 A
Power (P) = ?

P = VI

P = 110 × 0.5

P = 55 Watt

35. A bulb rated 5 volt – 100mA, calculate its (i) power (ii) resistance

Solution

Given values Voltage (V) = 5 volt
Current (I) = 100mA = \( 100 \times 10^{-3} \) A

P = VI

P = \( 5 \times 100 \times 10^{-3} \)

P = 0.5 Watt

36. Two electric bulbs rated 60W, 220V and 100W, 220 V. which one of them has higher resistance?

Solution

Given values (i) P1 = 60 W, V = 220 volt
(ii) P2 = 110 W , 220 volt
R1 = ?
R2 = ?

$$ R_1 = \frac{V^2}{P_1} $$

$$ R_1 = \frac{(220)^2}{60} $$

$$ R_1 = 806.67 \text{ Ohm} $$

$$ R_2 = \frac{V^2}{P_2} $$

$$ R_2 = \frac{(220)^2}{100} $$

$$ R_2 = 484 \text{ Ohm} $$

60 watt bulb has higher resistance than 100 watt bulb.

37. Two lamps rated 100 W, 220 Volt and 60W, 220 V connected in parallel. Calculate the current drawn by the circuit.

Solution

Given values P1 = 100W
P2 = 60 W
V = 220volt
I = ?

$$ I = \frac{V}{R} $$

In the above formula we don’t have value of R so we will have to find the value of R

$$ R_1 = \frac{V^2}{P_1} = \frac{(220)^2}{100} \text{ Ohm} $$

$$ R_2 = \frac{V^2}{P_2} = \frac{(220)^2}{60} \text{ Ohm} $$

Both the lamps are in parallel, so the resultant resistance is

$$ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} $$

$$ \frac{1}{R} = \frac{100}{220 \times 220} + \frac{60}{220 \times 220} $$

$$ \frac{1}{R} = \frac{100+60}{220 \times 220} $$

$$ \frac{1}{R} = \frac{160}{220 \times 220} \text{ Ohm} $$

Now we can find the value of current (I)

$$ I = \frac{220}{\frac{220 \times 220}{160}} $$

$$ I = \frac{220 \times 160}{220 \times 220} $$

$$ I = \frac{160}{220} = 0.72 A $$

38. An electric lamp is rated 100W, 220V. it is used for 10 hours daily then calculate energy consumed in KWh per day.

Solution

Given values P = 100 W
V = 220 Volt

$$ \text{Energy consumed per day} = \frac{P \times t}{1000} $$

$$ = \frac{100 \times 10}{1000} $$

Energy consumed per day = 1KWh or 1 Unit

39. A bulb rated 2.5 V and 650mA. Calculate (i) power (ii) resistance (iii) energy consumed for 5 hours.

Solution

Given values V = 2.5 Volt
I = 650 mA = 0.65A
T = 5 hours

(i) P=VI

P = 2.5 × 0.65 = 1.625 W

(ii) Resistance

$$ R = \frac{V}{I} $$

$$ R = \frac{2.5}{0.65} = 3.84 \text{ Ohm} $$

(iii) Energy consumed in 5 hours

$$ = \frac{P \times t}{1000} $$

$$ = \frac{1.625 \times 5}{1000} = 0.008125 \text{ KWh} $$

40. An immersion rod of 750W is used for one hour. Find the energy consumed in (i) KWh (ii) Joule

Solution

Given values P = 750 W
T = 1 hour

(i) Energy consumed in KWh

$$ \text{Energy consumed for 1 hour} = \frac{P \times t}{1000} $$

$$ = \frac{750 \times 1}{1000} = 0.75 \text{ KWh} $$

(iii) Energy consumed in Joule

1 KWh = \( 3.6 \times 10^6 \) Joule

So \( 0.75 \times 3.6 \times 10^6 \) Joule

= \( 2.7 \times 10^6 \) Joule

41. One heater is rated 220 V- 5 A. calculate the energy consumed if it is used for 5 hours.

Solution

Given values V = 220 volt
I = 5 A

$$ \text{Energy consumed for 5 hour} = \frac{P \times t}{1000} $$

But in this formula we need value of Power (P)

P = VI

P = 220 × 5

P = 1100W

Now we can put the value of P in the formula

$$ \text{Energy consumed for 5 hour} = \frac{1100 \times 5}{1000} $$

= 5.5 KWh

42. Which uses more energy a 250W computer in 1 hour or a 1200W heater in 10 minutes.

Solution

Given values P1 = 250W
T1 = 1hour
P2 = 1200W
T2 = 10 min = 10/60 hour

(i) A 250 W computer is used for 1 hour

= P1 × t1 = 250 × 1

= 250Wh

(ii) A 1200W heater is used for 10 min

= P2 × t2 = 1200 × 10/60

= 200Wh

So, computer uses more energy.

43. An electric heater is rated 1kW, 220V. calculate its resistance.

Solution

Given values P = 1kW = 1000W
V = 220 volt
R = ?

$$ R = \frac{V^2}{P} $$

$$ R = \frac{220^2}{1000} $$

R = 48.4 Ohm

44. An electric bulb rated 220 V- 100 W. calculate its resistance.

Solution

Given values Power (P) = 100W
Voltage (V) = 220volt
Resistance (R) = ?

$$ P = \frac{V^2}{R} $$

$$ 100 = \frac{220^2}{R} $$

$$ R = \frac{220^2}{100} $$

R = 484 Ohm

Numerical Problems based on Heating effect of current

45. How much heat will a device of 15 W produces in one minute if it is connected to a battery of 15 Volt.

Solution

Given values P = 15W
V = 15 volt
T = 1 min = 60 s

H = P × t

H = 15 × 60 = 900 J

46. 1 kJ hat is produced each second in 50 Ohm resistor. Calculate the potential difference across the resistor.

Solution

Given values H = 1kJ = 1000J
Time (t) = 1 s
Resistance (R) = 50 Ohm
Voltage (V) = ?

$$ H = I^2Rt $$

$$ 1000 = I^2 \times 50 \times 1 $$

$$ I^2 = \frac{1000}{50} = 20 $$

$$ I = \sqrt{20} = 4.47A $$

So, potential difference across the resistor

V = IR

V = 4.47 × 50 = 223.5 volt

47. A heater of resistance 10 Ohm draws 10 A current from electric source for 2 hours. Calculate the heat produced in the heater.

Solution

Given values Resistance (R) = 10 Ohm
Current (I) = 10 A
Time (t) = 2 hours = 2 × 60 × 60 = 7200 s
Heat (H) = ?

$$ H = I^2Rt $$

$$ H = (10)^2 \times 10 \times 7200 $$

H = \( 72 \times 10^5 \) J

48. 100 Joule of heat is produced each second in 10 Ohm resistance. find the potential difference across the resistor.

Solution

Given values Heat (H) = 100J
Time (t) = 1s
Resistance (R) = 10 Ohm
Potential difference (V) = ?

V = IR

In the above formula we need the value of current (I) so at first we must find the value of I form the given values

$$ H = I^2Rt $$

$$ I = \sqrt{\frac{H}{Rt}} $$

$$ I = \sqrt{\frac{100}{10 \times 1}} $$

$$ I = \sqrt{10} $$

$$ I = 3.16 A $$

Now we can find the value of potential difference

V = IR

V = 3.16 × 10

V = 31.6 volt

49. Find the heat produced in the following combination of resistors if current drawn for 10 seconds.

[Insert Image from Page 47 here: Two resistors in series (2Ω and 2Ω) connected to 10V]

Solution

Given values R1 = 2 Ohm
R2 = 2 Ohm
Potential difference (V) = 10 volt
Time (t) = 10 s
Heat (H) = ?

H = V I t

We will have to find the value of current (I)

R = R1 + R2

R = 2 + 2 = 4 Ohm

$$ I = \frac{V}{R} $$

$$ I = \frac{10}{4} = 2.5 A $$

H = V I t

H = 10 × 2.5 × 10

H = 250 J

50. How much energy is given to 10 coulomb charge passing through 15 volt battery?

Solution

Given values Charge (Q) = 10 Coulomb
Voltage (V) = 15 volt
Energy (E) = ?

E = VQ

E = 15 × 10 = 150 J

Solved Problems: Atoms and Molecules | Chemistry & Science Guide

Solved Problems: Atoms and Molecules

Table of Contents

1. Calculation of molar mass
2. Calculation based on number of moles from mass and volume
3. Calculation of mass from mole
4. Calculation based on number of atoms/molecules
5. Calculation based on molar volume
6. Calculation based on % composition

I. Calculation of molar mass

Calculate the gram molar mass of the following.

1) H₂O        2) CO₂        3) Ca₃(PO₄)₂

Solution:

1) H₂O

Atomic masses of H = 1, O = 16

Gram molar mass of H₂O = (1 × 2) + (16 × 1)

= 2 + 16

Gram molar mass of H₂O = 18 g

2) CO₂

Atomic masses of C = 12, O = 16

Gram molar mass of CO₂ = (12 × 1) + (16 × 2)

= 12 + 32

Gram molar mass of CO₂ = 44 g

3) Ca₃(PO₄)₂

Atomic masses of Ca = 40, P = 30, O = 16.

Gram molar mass of Ca₃(PO₄)₂ = (40 × 3) + [30 + (16 × 4)] × 2

= 120 + (94 × 2)

= 120 + 188

Gram molar mass of Ca₃(PO₄)₂ = 308 g

II. Calculation based on number of moles from mass and volume

1) Calculate the number of moles in 46 g of sodium?

= 46 / 23

2 moles of sodium

2) 5.6 litre of oxygen at S.T.P

3) Calculate the number of moles of a sample that contains 12.046 × 10²³ atoms of iron ?

III. Calculation of mass from mole

Calculate the mass of the following

1) 0.3 mole of aluminium (Atomic mass of Al = 27)

Mass = No. of moles × atomic mass

So, mass of Al = 0.3 × 27

= 8.1 g

2) 2.24 litre of SO₂ gas at S.T.P

Molecular mass of SO₂ = 32 + (16 × 2) = 32 + 32 = 64

Mass = No. of moles × molecular mass

Mass = 0.1 × 64

Mass of SO₂ = 6.4 g

3) 1.51 × 10²³ molecules of water

Molecular mass of H₂O = 18

0.25 = mass / 18

Mass = 0.25 × 18

Mass = 4.5 g

4) 5 × 10²³ molecules of glucose ?

Molecular mass of glucose = 180

= (180 × 5 × 10²³) / 6.023 × 10²³

= 149.43 g

IV. Calculation based on number of atoms/ molecules.

1) Calculate the number of molecules in 11.2 litre of CO₂ at S.T.P

= 11.2 / 22.4

= 0.5 mole

Number of molecules of CO₂ = number of moles of CO₂ × Avogadro’s number

= 0.5 × 6.023 × 10²³

= 3.011 × 10²³ molecules of CO₂

2) Calculate the number of atoms present in 1 gram of gold (Atomic mass of Au = 198)

Number of atoms of Au = 3.042 × 10²¹ g

3) Calculate the number of molecules in 54 gm of H₂O?

Number of molecules of water = 6.023 × 10²³ × 54 / 18

= 18.069 × 10²³ molecules

4) Calculate the number of atoms of oxygen and carbon in 5 moles of CO₂.

1 mole of CO₂ contains 2 moles of oxygen

5 moles of CO₂ contain 10 moles of oxygen

Number of atoms of oxygen = Number of moles of oxygen × Avogadro’s number

= 10 × 6.023 × 10²³

= 6.023 × 10²⁴ atoms of Oxygen

1 mole of CO₂ contains 1 mole of carbon

5 moles of CO₂ contains 5 moles of carbon

No. of atoms of carbon = No.of moles of carbon × Avogadro’s number

= 5 × 6.023 × 10²³

= 3.011 × 10²⁴ atoms of Carbon

V. Calculation based on molar volume

Calculate the volume occupied by:

1) 2.5 mole of CO₂ at S.T.P

2) 3.011 × 10²³ of ammonia gas molecules

= 2 moles

Volume occupied by NH₃ = number of moles × molar volume

= 2 × 22.4

= 44.8 litres at S.T.P

3) 14 g nitrogen gas

Number of moles = 14 / 28

= 0.5 mole

Volume occupied by N₂ at S.T.P = no. of moles × molar volume

0.5 × 22.4

11.2 litres.

VI. Calculation based on % composition

Calculate % of S in H₂SO₄

Molar mass of H₂SO₄ = (1 × 2) + (32 × 1) + (16 × 4)

= 2 + 32 + 64

= 98 g