Electricity class 10 numericals with all formulas of electricity class 10 with descriptive. Solve the most important 50 numericals on electricity for class 10 with solutions by clicking on the show explanation.
Electricity Numericals Class 10
Comprehensive Solved Problems with Step-by-Step Explanations
Key Concepts & Formulas
Solved Numerical Questions
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1. Calculate the current through a $10\ \Omega$ resistor when connected to a $5\ V$ battery.
Given: $V = 5\ V, R = 10\ \Omega$
Using Ohm’s Law: $$I = \frac{V}{R} = \frac{5}{10} = 0.5\ A$$
Answer: The current is $0.5\ A$.
2. A $60\ W$ bulb is connected to a $220\ V$ supply. Calculate the current.
Given: $P = 60\ W, V = 220\ V$
Using Power formula: $$I = \frac{P}{V} = \frac{60}{220} \approx 0.273\ A$$
3. Find the equivalent resistance of $5\ \Omega, 10\ \Omega,$ and $15\ \Omega$ in series.
For series connection: $$R_{eq} = R_1 + R_2 + R_3$$
$$R_{eq} = 5 + 10 + 15 = 30\ \Omega$$
4. A $12\ V$ battery is connected across a $4\ \Omega$ resistor. Calculate current.
$$I = \frac{V}{R} = \frac{12}{4} = 3\ A$$
5. Determine potential difference across a $20\ \Omega$ resistor with $0.5\ A$ current.
$$V = IR = 0.5 \times 20 = 10\ V$$
6. Calculate equivalent resistance of $6\ \Omega$ and $12\ \Omega$ in parallel.
Formula: $$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$$
$$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$$
$$R_p = 4\ \Omega$$
7. Heater rated $1500\ W, 220\ V$. Calculate resistance.
$$R = \frac{V^2}{P} = \frac{220^2}{1500} = \frac{48400}{1500} \approx 32.27\ \Omega$$
8. $3\ \Omega$ in series with parallel combo of $6\ \Omega$ and $12\ \Omega$. Find total R.
Parallel part ($R_p$) of $6\ \Omega$ and $12\ \Omega$ is $4\ \Omega$.
Total Series: $$R_{total} = 3 + R_p = 3 + 4 = 7\ \Omega$$
9. $2000\ W$ kettle runs for 3 hours. Energy in kWh?
$$P = 2000\ W = 2\ kW$$
$$E = P \times t = 2\ kW \times 3\ h = 6\ kWh$$
10. $R=10\ \Omega, I=2\ A$. Heat produced in 5 mins?
$t = 5 \times 60 = 300\ s$
$$H = I^2Rt = (2)^2 \times 10 \times 300 = 4 \times 3000 = 12,000\ J$$
11. Find resistivity: $R=5\ \Omega, L=2\ m, A=0.01\ m^2$.
$$\rho = \frac{R \cdot A}{L} = \frac{5 \times 0.01}{2} = 0.025\ \Omega m$$
12. Device draws $0.2\ A$ at $12\ V$. Calculate power.
$$P = VI = 12 \times 0.2 = 2.4\ W$$
13. $4\ \Omega$ and $8\ \Omega$ in parallel. $V=12\ V$. Find total Current.
$$R_p = \frac{4 \times 8}{4 + 8} = \frac{32}{12} = \frac{8}{3}\ \Omega$$
$$I = \frac{V}{R_p} = \frac{12}{8/3} = \frac{36}{8} = 4.5\ A$$
14. Time taken by $100\ W$ bulb to consume $1\ kWh$.
$$t = \frac{E}{P} = \frac{1000\ Wh}{100\ W} = 10\ \text{hours}$$
15. Wire $R=2\ \Omega$ is stretched to double length. New Resistance?
When length doubles ($2L$), Area halves ($A/2$).
$$R_{new} = \rho \frac{2L}{A/2} = 4 \left( \rho \frac{L}{A} \right) = 4R$$
$$R_{new} = 4 \times 2 = 8\ \Omega$$
16. Find charge: $I = 3\ A$, time $= 2$ minutes.
$$t = 120\ s$$
$$Q = It = 3 \times 120 = 360\ C$$
17. $60\ W$ lamp used 5 hours daily for 30 days. Total energy?
Daily: $$0.06\ kW \times 5\ h = 0.3\ kWh$$
Total: $$0.3 \times 30 = 9\ kWh$$
18. $R=10\ \Omega$ at $20^\circ C$, $\alpha=0.004/^\circ C$. Find $R$ at $50^\circ C$.
$$R_t = R_0[1 + \alpha \Delta T]$$
$$R_{50} = 10[1 + 0.004(30)] = 10[1.12] = 11.2\ \Omega$$
19. $100\ \Omega$ resistor on $20\ V$ battery. Power dissipated?
$$P = \frac{V^2}{R} = \frac{20^2}{100} = \frac{400}{100} = 4\ W$$
20. Charge of $50\ C$ flows in 10 seconds. Calculate Current.
$$I = \frac{Q}{t} = \frac{50}{10} = 5\ A$$
21. Resistor $8\ \Omega$, Capacitor $0.5\ F$ on $10\ V$. Charge stored?
In DC steady state, capacitor fully charges to source voltage.
$$Q = CV = 0.5 \times 10 = 5\ C$$
22. Energy stored in capacitor $10\ \mu F$ at $50\ V$.
$$E = \frac{1}{2}CV^2$$
$$E = 0.5 \times (10 \times 10^{-6}) \times 2500 = 0.0125\ J$$
23. Wire $10\ \Omega$ stretched to double length. New R?
As per stretching rule ($R \propto L^2$):
$$R’ = n^2 R = 2^2 \times 10 = 40\ \Omega$$
24. Iron consumes $1\ kWh$ in 2 hours. Power rating?
$$P = \frac{E}{t} = \frac{1000\ Wh}{2\ h} = 500\ W$$
25. $L=1\ m, A=0.01\ m^2, R=2\ \Omega$. Find Resistivity.
$$\rho = \frac{RA}{L} = \frac{2 \times 0.01}{1} = 0.02\ \Omega m$$
26. $I=2\ A, R=3\ \Omega$, time 10 mins. Heat generated?
$$t = 600\ s$$
$$H = I^2Rt = 2^2 \times 3 \times 600 = 7200\ J$$
27. $100\ W$ fan used 5 hrs daily for 15 days. Energy?
Daily: $0.1\ kW \times 5 = 0.5\ kWh$
Total: $0.5 \times 15 = 7.5\ kWh$
28. $\rho=1.7 \times 10^{-8}\ \Omega m, L=2\ m$, dia $1\ mm$. Find R.
Radius $r = 0.5\ mm = 5 \times 10^{-4}\ m$
Area $A = \pi r^2 \approx 7.85 \times 10^{-7}\ m^2$
$$R = \frac{\rho L}{A} = \frac{1.7 \times 10^{-8} \times 2}{7.85 \times 10^{-7}} \approx 0.043\ \Omega$$
29. Total $R=15\ \Omega, V=45\ V$. Calculate Current.
$$I = \frac{45}{15} = 3\ A$$
30. Bulb rated $100\ W, 220\ V$. Find Resistance.
$$R = \frac{220^2}{100} = 484\ \Omega$$
31. $2\ \Omega, 4\ \Omega, 6\ \Omega$ in parallel. Equivalent R?
$$\frac{1}{R} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} = \frac{6+3+2}{12} = \frac{11}{12}$$
$$R = \frac{12}{11} \approx 1.09\ \Omega$$
32. Resistor dissipates $50\ W$ at $10\ V$. Find Current.
$$I = \frac{P}{V} = \frac{50}{10} = 5\ A$$
33. Motor takes $10\ A$ from $220\ V$. Power consumed?
$$P = VI = 220 \times 10 = 2200\ W$$
34. $24\ V$ battery, resistors $2, 4, 6\ \Omega$ in series. Current?
$$R_{total} = 2 + 4 + 6 = 12\ \Omega$$
$$I = \frac{24}{12} = 2\ A$$
35. Current $0.5\ A$ at $240\ V$. Calculate Power.
$$P = 240 \times 0.5 = 120\ W$$
36. $60\ C$ charge in 15 seconds. Current?
$$I = \frac{60}{15} = 4\ A$$
37. Wire $5\ \Omega$ bent into circle. Resistance across diameter?
The wire is split into two semicircles of $2.5\ \Omega$ each, connected in parallel.
$$R_{eq} = \frac{2.5}{2} = 1.25\ \Omega$$
38. Bulb $100\ W$ at $250\ V$. Resistance and Current?
$$R = \frac{250^2}{100} = 625\ \Omega$$
$$I = \frac{100}{250} = 0.4\ A$$
39. $40\ W$ lamp at $220\ V$. Resistance and Current?
$$R = \frac{220^2}{40} = 1210\ \Omega$$
$$I = \frac{40}{220} \approx 0.18\ A$$
40. $9\ V$ battery on $3\ \Omega$ resistor. Power?
$$P = \frac{9^2}{3} = \frac{81}{3} = 27\ W$$
41. $12\ V$ battery. Circuit: $4\ \Omega || 6\ \Omega$ then series with $3\ \Omega$.
Parallel: $$R_p = \frac{4 \times 6}{10} = 2.4\ \Omega$$
Total R: $$2.4 + 3 = 5.4\ \Omega$$
Current: $$I = \frac{12}{5.4} \approx 2.22\ A$$
42. Copper wire temp change $20^\circ C$, $\alpha=0.004$. Initial $R=0.0255\ \Omega$.
$$R_{new} = 0.0255[1 + 0.004(20)] = 0.0255(1.08) \approx 0.0275\ \Omega$$
43. $1500\ W$ kettle, 3 hours. Cost at $0.12 per kWh?
$$E = 1.5\ kW \times 3\ h = 4.5\ kWh$$
$$Cost = 4.5 \times 0.12 = \$0.54$$
44. $60\ W$ and $100\ W$ bulbs in series to $220\ V$. Total Power?
In series, power adds reciprocally: $$\frac{1}{P_{total}} = \frac{1}{P_1} + \frac{1}{P_2}$$
$$\frac{1}{P} = \frac{1}{60} + \frac{1}{100} = \frac{160}{6000}$$
$$P_{total} = 37.5\ W$$
45. Transformer $P_{in}=1000\ W$, $V_{sec}=120\ V$, Efficiency 90%. $I_{sec}$?
$$P_{out} = 0.9 \times 1000 = 900\ W$$
$$I_{sec} = \frac{900}{120} = 7.5\ A$$
46. Battery EMF $12\ V$, internal res $0.5\ \Omega$, load $5\ \Omega$. Terminal V?
$$I = \frac{12}{5 + 0.5} = \frac{12}{5.5} \approx 2.18\ A$$
$$V_{term} = E – Ir = 12 – (2.18 \times 0.5) \approx 10.91\ V$$
47. $6\ \mu F$ and $12\ \mu F$ capacitors in series. Equivalent C?
Capacitors in series add like resistors in parallel.
$$\frac{1}{C} = \frac{1}{6} + \frac{1}{12} = \frac{3}{12}$$
$$C_{eq} = 4\ \mu F$$
48. $2000\ W$ heater on $230\ V$. Current and Resistance?
$$I = \frac{2000}{230} \approx 8.7\ A$$
$$R = \frac{230}{8.7} \approx 26.44\ \Omega$$
49. $R$ goes from $5\ \Omega$ to $5.5\ \Omega$ over $80^\circ C$. Find $\alpha$.
$$\alpha = \frac{\Delta R}{R_0 \Delta T} = \frac{0.5}{5 \times 80} = \frac{0.5}{400} = 0.00125\ /^\circ C$$
50. Wheatstone Bridge: $10, 20, 30, 40\ \Omega$. Balanced?
Condition: $$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$
$$\frac{10}{20} = 0.5$$ vs $$\frac{30}{40} = 0.75$$
Answer: Not balanced.