Current Electricity Class 9 ICSE Important Questions with Answers

Answer: (b) ampere.

Current is charge per unit time, $I=\dfrac{Q}{t}$I=Q/t, measured in ampere $(\text{A})$(A), where $1\ \text{A}=1\ \text{C s}^{-1}$1 A=1 C s^-1.

Answer: (b) from the negative terminal to the positive terminal.

Electrons are negatively charged, so in the external circuit they move from the negative terminal towards the positive terminal. The conventional current is taken in the opposite direction, from the positive terminal to the negative terminal.

Answer: (c) graphite.

Graphite (a form of carbon) conducts electricity, which is why it is used for the electrodes of dry cells. Glass, dry wood and rubber are insulators: they do not allow charge to flow through them easily.

Answer: (b) $10^{19}$10¹⁹.

In $1$1 s the charge that flows is $Q=It=1.6\times1=1.6\ \text{C}$Q=It=1.6×1=1.6 C. Number of electrons $n=\dfrac{Q}{e}=\dfrac{1.6}{1.6\times10^{-19}}=10^{19}$n=Q/e=1.61.6×10^-19=10¹⁹.

Answer: (d) superconductor. Good conductors such as copper have low (but not zero) resistance; semiconductors such as silicon lie between conductors and insulators; superconductors have zero resistance only when cooled below a critical temperature.

Answer: (a) Both A and R are true and R is the correct explanation of A. Current flows only in a closed circuit, a complete conducting path from one terminal of the cell back to the other. Opening the key makes it an open circuit, so no current flows and the bulb does not glow.

(a) Electric current: the rate of flow of electric charge through a conductor, $I=\dfrac{Q}{t}$I=Q/t. SI unit: ampere $(\text{A})$(A).

(b) Potential difference: the work done in moving unit positive charge from one point to another, $V=\dfrac{W}{Q}$V=W/Q. SI unit: volt $(\text{V})$(V).

Given: $Q=60\ \text{C}$Q=60 C, $t=2\ \text{min}=2\times60=120\ \text{s}$t=2 min=2×60=120 s.

$I=\frac{Q}{t}=\frac{60}{120}=0.5\ \text{A}.$I=Q/t=60/120=0.5 A.

The current in the conductor is $0.5\ \text{A}$0.5 A.

Charge is quantised: the charge on any body is always a whole-number multiple of the electronic charge, $q=ne$q=ne, where $n=1,2,3,\dots$n=1,2,3,…, because a body can gain or lose only whole electrons.

$\dfrac{2.4\times10^{-19}}{1.6\times10^{-19}}=1.5$2.4×10^-191.6×10^-19=1.5, which is not a whole number, so no, a body cannot carry this charge.

Ohm's law: the current through a conductor is directly proportional to the potential difference across its ends, provided its temperature and other physical conditions stay the same: $V\propto I$V∝ I, so $V=IR$V=IR.

Resistance $R$R is the opposition a conductor offers to the flow of current, $R=\dfrac{V}{I}$R=V/I. Its SI unit is the ohm ($\Omega$Ω): a conductor has a resistance of $1\ \Omega$1 Ω if a current of $1\text{ A}$1 A flows through it when the potential difference across it is $1\text{ V}$1 V.

Potential difference $V=\dfrac{W}{Q}=\dfrac{45}{15}=3\ \text{V}$V=W/Q=45/15=3 V.

For $5\ \text{C}$5 C between the same points: $W=VQ=3\times5=15\ \text{J}$W=VQ=3×5=15 J.

(A potential difference of $3\ \text{V}$3 V means $3\ \text{J}$3 J of work is done for every coulomb of charge moved.)

In the cell symbol, the longer, thinner line is the positive terminal and the shorter, thicker line is the negative terminal. A galvanometer is drawn like the ammeter, with G in place of A.

Conductors allow electric charge to flow through them easily, because they have many free electrons. Examples: copper, aluminium, silver, iron, graphite.

Insulators do not allow charge to flow through them easily, because they have almost no free electrons. Examples: rubber, plastic, glass, dry wood.

Copper is a good conductor, so it carries the current. The plastic covering is an insulator: it stops the current from reaching anyone who touches the wire (preventing electric shock) and stops bare wires touching each other (preventing a short circuit).

(i) $R=\dfrac{V}{I}=\dfrac{6}{0.5}=\boxed{12\ \Omega}$R=V/I=6/0.5=12 Ω.

(ii) $I=\dfrac{V}{R}=\dfrac{3}{12}=\boxed{0.25\text{ A}}$I=V/R=3/12=0.25 A (half the voltage, half the current).

(iii) $V=IR=2\times4=\boxed{8\text{ V}}$V=IR=2×4=8 V.

(a)

Conventional current leaves the positive terminal of the cell (the longer line), flows round the circuit and returns to the negative terminal.

(b) The key closes or opens the circuit, switching the current on or off. The ammeter, connected in series, detects and measures the current. The rheostat is a variable resistance: sliding its contact changes the current in the circuit, so the bulb becomes brighter or dimmer.

(c) Opening the key makes it an open circuit, so no current flows and the ammeter reads zero (the bulb goes off).

Given $I=0.5\ \text{A}$I=0.5 A, $t=10\ \text{min}=600\ \text{s}$t=10 min=600 s, $V=6\ \text{V}$V=6 V.

(a) $Q=It=0.5\times600=300\ \text{C}$Q=It=0.5×600=300 C.

(b) $n=\dfrac{Q}{e}=\dfrac{300}{1.6\times10^{-19}}=1.875\times10^{21}$n=Q/e=3001.6×10^-19=1.875×10²¹ electrons.

(c) From $V=\dfrac{W}{Q}$V=W/Q: $W=VQ=6\times300=1800\ \text{J}$W=VQ=6×300=1800 J.

(i) Cells supply direct current (d.c.), which flows in one direction only. Another source of d.c. is an accumulator (a rechargeable storage cell, such as a car battery).

(ii) $t=5\ \text{min}=300\ \text{s}$t=5 min=300 s, so $Q=It=0.3\times300=90\ \text{C}$Q=It=0.3×300=90 C.

(iii) $W=VQ=3\times90=270\ \text{J}$W=VQ=3×90=270 J.

(iv) By Ohm's law, $R=\dfrac{V}{I}=\dfrac{3}{0.3}=10\ \Omega$R=V/I=3/0.3=10 Ω.