Electricity — Important Questions

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

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

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

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

Answer: (a) Both A and R are true and R is the correct explanation of A - nichrome has a high resistivity, so it produces a large amount of heat ($H=I^2Rt$) for a given current, and its high melting point lets it glow red-hot without melting or oxidising readily.

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

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

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

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

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

(a) See the diagram: the three resistors are connected between the same two points (across the $12\ \text{V}$ battery).
(b) (i) Series: $R_s=6+6+3=15\ \Omega$.
(ii) Parallel: $\frac{1}{R_p}=\frac{1}{6}+\frac{1}{6}+\frac{1}{3}=\frac{1}{6}+\frac{1}{6}+\frac{2}{6}=\frac{4}{6}=\frac{2}{3}$, so $R_p=1.5\ \Omega$.
(c) The series combination ($15\ \Omega$) has the greater resistance.
(d) For the parallel combination, $I=\frac{V}{R_p}=\frac{12}{1.5}=8\ \text{A}$.

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

(i) In parallel each appliance gets the full $220\ \text{V}$, 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.