Current Electricity — ICSE Class 9 Physics Important Questions

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) $V=IR$V=IR.

Ohm's law states that at constant temperature the current through a conductor is directly proportional to the potential difference, giving $V=IR$V=IR.

Answer: (b) $2\ \Omega$2.

$\dfrac{1}{R_p}=\dfrac{1}{3}+\dfrac{1}{6}=\dfrac{2+1}{6}=\dfrac{3}{6}=\dfrac{1}{2}$1/R_p=1/3+1/6=2+1/6=3/6=1/2, so $R_p=2\ \Omega$R_p=2.

Answer: (c) $4R$4R.

When length doubles, the area halves (constant volume). Since $R\propto\dfrac{l}{A}$R/A, doubling $l$l and halving $A$A multiplies resistance by $2\times2=4$2×2=4, giving $4R$4R.

Answer: (a) A voltmeter is connected in parallel so that it reads the full potential difference across the component; its very high resistance ensures it draws negligible current and does not disturb the circuit. R correctly explains A.

(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.

Ohm's law: At constant temperature, the current $I$I through a conductor is directly proportional to the potential difference $V$V across its ends, so $V=IR$V=IR.

Resistance:
$R=\frac{V}{I}=\frac{8}{0.4}=20\ \Omega.$R=V/I=8/0.4=20 .

Current when voltage is doubled to $16\ \text{V}$16 V (resistance unchanged):
$I=\frac{V}{R}=\frac{16}{20}=0.8\ \text{A}.$I=V/R=16/20=0.8 A.

The resistance is $20\ \Omega$20 and the current doubles to $0.8\ \text{A}$0.8 A.

The resistance $R$R of a wire depends on:

1. Length $(l)$(l): $R\propto l$R l; a longer wire has more resistance.
2. Area of cross-section $(A)$(A): $R\propto\dfrac{1}{A}$R1/A; a thicker wire has less resistance.
3. Nature of the material (resistivity $\rho$): different materials have different resistivities.
4. Temperature: the resistance of a metallic conductor increases as its temperature rises.

Combining the first three, $R=\rho\dfrac{l}{A}$R=/A.

(a) Total resistance (series):
$R_s=2+3+5=10\ \Omega.$R_s=2+3+5=10 .

(b) Current in the circuit:
$I=\frac{V}{R_s}=\frac{10}{10}=1\ \text{A}.$I=V/R_s=10/10=1 A.
(The current is the same through each resistor in series.)

(c) P.d. across the $3\ \Omega$3 resistor:
$V_3=I\times R_3=1\times3=3\ \text{V}.$V_3=I× R_3=1×3=3 V.

(a) The ammeter (A) is in series with the resistor (R) and the voltmeter (V) is across the resistor.

(b) The ammeter is connected in series (so the full circuit current passes through it) and the voltmeter is connected in parallel across the resistor (to read the p.d. across it).

(c) $R=\dfrac{V}{I}=\dfrac{6}{2}=3\ \Omega.$R=V/I=6/2=3 .

(a) Parallel combination:
$\frac{1}{R_p}=\frac{1}{4}+\frac{1}{12}=\frac{3+1}{12}=\frac{4}{12}=\frac{1}{3}\ \Rightarrow\ R_p=3\ \Omega.$1/R_p=1/4+1/12=3+1/12=4/12=1/3 R_p=3 .

(b) Total resistance (parallel part in series with $2\ \Omega$2):
$R=R_p+2=3+2=5\ \Omega.$R=R_p+2=3+2=5 .

(c) Current from the battery:
$I=\frac{V}{R}=\frac{12}{5}=2.4\ \text{A}.$I=V/R=12/5=2.4 A.

(d) P.d. across the parallel combination $=I\times R_p=2.4\times3=7.2\ \text{V}$=I× R_p=2.4×3=7.2 V. This is the voltage across the $12\ \Omega$12 resistor, so
$I_{12}=\frac{7.2}{12}=0.6\ \text{A}.$I_12=7.2/12=0.6 A.

The current through the $12\ \Omega$12 resistor is $0.6\ \text{A}$0.6 A.

(i) By Ohm's law, $V=IR$V=IR, so $R=\dfrac{V}{I}$R=V/I.

(ii) $R=\dfrac{V}{I}=\dfrac{6}{1.5}=4\ \Omega.$R=V/I=6/1.5=4 .

(iii) In series the total resistance $=R+R=4+4=8\ \Omega$=R+R=4+4=8, so
$I=\frac{V}{R_s}=\frac{6}{8}=0.75\ \text{A}.$I=V/R_s=6/8=0.75 A.
The ammeter now reads $0.75\ \text{A}$0.75 A.

(iv) In parallel:
$\frac{1}{R_p}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\ \Rightarrow\ R_p=2\ \Omega.$1/R_p=1/4+1/4=2/4=1/2 R_p=2 .
The total resistance is $2\ \Omega$2.