Electricity — CBSE Class 10 Science Important Questions

Answer: (a) Ohm ($\Omega$Ω) - from Ohm's law $R=\frac{V}{I}$R=V/I, so $1\ \Omega=1\ \text{V/A}$1 Ω=1 V/A.

Answer: (a) $10\ \Omega$10 Ω - in series $R_s=R_1+R_2+R_3=2+3+5=10\ \Omega$R_s=R₁+R₂+R₃=2+3+5=10 Ω.

Answer: (a) $3\ \Omega$3 Ω - in parallel $\frac{1}{R_p}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}$1/R_p=1/6+1/6=2/6, so $R_p=3\ \Omega$R_p=3 Ω.

Answer: (a) $484\ \Omega$484 Ω - using $P=\frac{V^2}{R}$P=V²/R, we get $R=\frac{V^2}{P}=\frac{(220)^2}{100}=\frac{48400}{100}=484\ \Omega$R=V²/P=(220)²/100=48400/100=484 Ω.

Answer: (a) Both A and R are true and R is the correct explanation of A - from $V=\frac{W}{Q}$V=W/Q, the work done is $W=VQ=12\times2=24\ \text{J}$W=VQ=12×2=24 J.

Electric current is the rate of flow of electric charge through a conductor. Its SI unit is the ampere (A). The relation is $I=\frac{Q}{t}$I=Q/t, where $Q$Q is the charge (in coulomb) flowing in time $t$t (in second); thus $1\ \text{A}=1\ \text{C/s}$1 A=1 C/s.

(i) The resistivity of an alloy is much higher than that of its constituent pure metals, so it produces more heat ($H=I^2Rt$H=I²Rt) for the same current. (ii) An alloy does not oxidise (burn) readily at high temperatures, so the element is not damaged even when it becomes red-hot.

(a) In series, $R_s=5+10+15=30\ \Omega$R_s=5+10+15=30 Ω.
(b) By Ohm's law, $I=\frac{V}{R_s}=\frac{6}{30}=0.2\ \text{A}$I=V/R_s=6/30=0.2 A.
(c) The same current flows through each resistor, so across the $15\ \Omega$15 Ω resistor $V=IR=0.2\times15=3\ \text{V}$V=IR=0.2×15=3 V.

(a) $\frac{1}{R_p}=\frac{1}{10}+\frac{1}{15}=\frac{3}{30}+\frac{2}{30}=\frac{5}{30}=\frac{1}{6}$1/R_p=1/10+1/15=3/30+2/30=5/30=1/6, so $R_p=6\ \Omega$R_p=6 Ω.
(b) Total current $I=\frac{V}{R_p}=\frac{6}{6}=1\ \text{A}$I=V/R_p=6/6=1 A.
(c) In parallel the full $6\ \text{V}$6 V acts across each resistor, so current through the $10\ \Omega$10 Ω resistor is $I_1=\frac{V}{R_1}=\frac{6}{10}=0.6\ \text{A}$I₁=V/R₁=6/10=0.6 A.

(a) Joule's law of heating: the heat produced in a resistor is directly proportional to the square of the current ($H \propto I^2$H ∝ I²), to the resistance ($H \propto R$H ∝ R) and to the time for which the current flows ($H \propto t$H ∝ t); that is, $H=I^2Rt$H=I²Rt.
(b) $H=I^2Rt=(5)^2\times20\times30=25\times20\times30=15000\ \text{J}$H=I²Rt=(5)²×20×30=25×20×30=15000 J.
(c) Tungsten has a very high melting point (about $3380^{\circ}\text{C}$3380°C) and high resistivity, so it can be heated to a high temperature and glow (emit light) without melting.

(a) Radius $r=0.25\ \text{mm}=2.5\times10^{-4}\ \text{m}$r=0.25 mm=2.5×10^-4 m, so the area of cross-section is $A=\pi r^2=3.14\times(2.5\times10^{-4})^2=1.9625\times10^{-7}\ \text{m}^2$A=π r²=3.14×(2.5×10^-4)²=1.9625×10^-7 m².
From $R=\rho\frac{l}{A}$R=ρ l/A: $l=\frac{RA}{\rho}=\frac{10\times1.9625\times10^{-7}}{1.6\times10^{-8}}\approx122.7\ \text{m}$l=RA/ρ=10×1.9625×10^-71.6×10^-8≈122.7 m.
(b) Doubling the diameter makes the area of cross-section $4$4 times as large. Since $R\propto\frac{1}{A}$R∝ 1/A, the resistance becomes $\frac{1}{4}$1/4 of $10\ \Omega$10 Ω, i.e. $2.5\ \Omega$2.5 Ω.

(a) The three resistors are connected between the same two junctions, across the $12\ \text{V}$12 V battery:

(b) (i) Series: $R_s=6+6+3=15\ \Omega$R_s=6+6+3=15 Ω.
(ii) Parallel: $\frac{1}{R_p}=\frac{1}{6}+\frac{1}{6}+\frac{1}{3}=\frac{1}{6}+\frac{1}{6}+\frac{2}{6}=\frac{4}{6}=\frac{2}{3}$1/R_p=1/6+1/6+1/3=1/6+1/6+2/6=4/6=2/3, so $R_p=1.5\ \Omega$R_p=1.5 Ω.
(c) The series combination ($15\ \Omega$15 Ω) has the greater resistance.
(d) For the parallel combination, $I=\frac{V}{R_p}=\frac{12}{1.5}=8\ \text{A}$I=V/R_p=12/1.5=8 A.

(a) Electrical power is $P=VI$P=VI. Using Ohm's law $V=IR$V=IR, substitute for $V$V: $P=(IR)\times I=I^2R$P=(IR)× I=I²R.
(b) Here $P=2000\ \text{W}$P=2000 W, $V=220\ \text{V}$V=220 V.
(i) $P=VI$P=VI, so $I=\frac{P}{V}=\frac{2000}{220}=9.09\ \text{A}$I=P/V=2000/220=9.09 A.
(ii) $P=\frac{V^2}{R}$P=V²/R, so $R=\frac{V^2}{P}=\frac{(220)^2}{2000}=\frac{48400}{2000}=24.2\ \Omega$R=V²/P=(220)²/2000=48400/2000=24.2 Ω.
(iii) Energy per day $=2\ \text{kW}\times3\ \text{h}=6\ \text{kWh}$=2 kW×3 h=6 kWh; for 30 days $=6\times30=180\ \text{kWh}$=6×30=180 kWh (units). Cost $=180\times5=\textbf{Rs }900$=180×5=Rs 900.

(i) In parallel each appliance gets the full $220\ \text{V}$220 V, can be switched on/off independently, and if one appliance stops working the others keep operating; each also draws its own required current.
(ii) A fuse is a safety device that melts and breaks the circuit when the current exceeds a safe value, protecting the wiring and appliances from damage due to overloading or a short circuit.
(iii) A fuse wire is made of a material with a low melting point and suitable resistivity (usually an alloy of tin and lead).
(iv) A thick copper wire has a high melting point and will not melt during an overload or short circuit, so the excessive current continues to flow and can cause overheating and fire - the very hazard the fuse is meant to prevent.