Current Electricity — ICSE Class 10 Physics Important Questions

Answer: (b) $\Omega\,\text{m}$\,m.

From $R=\dfrac{\rho l}{A}$R= l/A, $\rho=\dfrac{RA}{l}$=RA/l, whose unit is $\dfrac{\Omega\cdot\text{m}^2}{\text{m}}=\Omega\,\text{m}$·m^2/m=\,m.

Answer: (d) $2\,\Omega$2\,.

$\dfrac{1}{R_p}=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}$1/R_p=1/6+1/6+1/6=3/6=1/2, so $R_p=2\,\Omega$R_p=2\,. (For $n$n equal resistors in parallel, $R_p=\dfrac{R}{n}$R_p=R/n.)

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

On doubling the length at constant volume, the area halves. $R=\dfrac{\rho l}{A}$R= l/A, so $R\propto\dfrac{l}{A}$R/A. New $R'=\dfrac{\rho(2l)}{A/2}=4\dfrac{\rho l}{A}=4R$R'=(2l)/A/2=4 l/A=4R.

Answer: (b) $1.75\,\text{V}$1.75\,V.

Current $I=\dfrac{\varepsilon}{R+r}=\dfrac{2}{3.5+0.5}=\dfrac{2}{4}=0.5\,\text{A}$I=/R+r=2/3.5+0.5=2/4=0.5\,A. Terminal voltage $V=\varepsilon-Ir=2-(0.5\times0.5)=2-0.25=1.75\,\text{V}$V=-Ir=2-(0.5×0.5)=2-0.25=1.75\,V.

Answer: (a) Parallel branches give extra paths so the total current increases for the same voltage, which means the effective resistance falls below even the smallest branch resistance. R correctly explains A.

Ohm's law: At constant temperature, the electric current flowing through a conductor is directly proportional to the potential difference across its ends, i.e. $V\propto I$V I, or $V=IR$V=IR where $R$R is a constant called the resistance.

Quantity kept constant: the temperature (and physical conditions) of the conductor must remain constant.

The resistance $R=\dfrac{\rho l}{A}$R= l/A depends on:

(i) Length ($l$l): resistance is directly proportional to length - a longer wire has more resistance.

(ii) Area of cross-section ($A$A): resistance is inversely proportional to the area of cross-section - a thicker wire has less resistance.

(It also depends on the material's resistivity $\rho$ and on temperature.)

(i) Parallel: $\dfrac{1}{R_p}=\dfrac{1}{4}+\dfrac{1}{6}=\dfrac{3+2}{12}=\dfrac{5}{12}$1/R_p=1/4+1/6=3+2/12=5/12, so $R_p=\dfrac{12}{5}=2.4\,\Omega$R_p=12/5=2.4\,.

(ii) Total current $I=\dfrac{V}{R_p}=\dfrac{12}{2.4}=5\,\text{A}$I=V/R_p=12/2.4=5\,A.

(iii) In parallel the full $12\,\text{V}$12\,V acts across the $6\,\Omega$6\, resistor: $I_6=\dfrac{V}{R}=\dfrac{12}{6}=2\,\text{A}$I_6=V/R=12/6=2\,A.

Equivalent resistance $=2.4\,\Omega$=2.4\,, total current $=5\,\text{A}$=5\,A, current in the $6\,\Omega$6\, resistor $=2\,\text{A}$=2\,A.

Given $l=2\,\text{m}$l=2\,m, $A=0.5\,\text{mm}^2=0.5\times10^{-6}\,\text{m}^2$A=0.5\,mm^2=0.5×10^-6\,m^2, $\rho=1.7\times10^{-8}\,\Omega\,\text{m}$=1.7×10^-8\,\,m.

$R=\dfrac{\rho l}{A}=\dfrac{1.7\times10^{-8}\times2}{0.5\times10^{-6}}$R= l/A=1.7×10^-8×20.5×10^-6

$R=\dfrac{3.4\times10^{-8}}{0.5\times10^{-6}}=6.8\times10^{-2}\,\Omega=0.068\,\Omega$R=3.4×10^-80.5×10^-6=6.8×10^-2\,=0.068\,

The resistance of the wire is $0.068\,\Omega$0.068\,.

Parallel part: two $6\,\Omega$6\, resistors, $R_p=\dfrac{6\times6}{6+6}=\dfrac{36}{12}=3\,\Omega$R_p=6×6/6+6=36/12=3\,.

Total (series with the $3\,\Omega$3\,): $R=3+R_p=3+3=6\,\Omega$R=3+R_p=3+3=6\,.

Current from battery: $I=\dfrac{V}{R}=\dfrac{9}{6}=1.5\,\text{A}$I=V/R=9/6=1.5\,A.

The total resistance is $6\,\Omega$6\, and the battery supplies $1.5\,\text{A}$1.5\,A.

(a) The EMF of a cell is the total energy supplied by the cell per unit charge in driving the charge round the complete circuit (measured as the potential difference across its terminals when no current is drawn). The terminal voltage is the potential difference across the terminals of the cell when current is being drawn by an external circuit; it equals the potential difference across the external resistance.

(b) $\varepsilon=6\,\text{V}$=6\,V, $r=0.5\,\Omega$r=0.5\,, $R=11.5\,\Omega$R=11.5\,.

(i) $I=\dfrac{\varepsilon}{R+r}=\dfrac{6}{11.5+0.5}=\dfrac{6}{12}=0.5\,\text{A}$I=/R+r=6/11.5+0.5=6/12=0.5\,A.

(ii) Terminal voltage $V=IR=0.5\times11.5=5.75\,\text{V}$V=IR=0.5×11.5=5.75\,V (or $V=\varepsilon-Ir=6-0.25=5.75\,\text{V}$V=-Ir=6-0.25=5.75\,V).

(iii) Voltage drop across internal resistance $=Ir=0.5\times0.5=0.25\,\text{V}$=Ir=0.5×0.5=0.25\,V.

(Check: $5.75+0.25=6\,\text{V}=\varepsilon$5.75+0.25=6\,V=.)

(a) Connect the resistance wire (conductor) in series with a battery, a key, a rheostat and an ammeter, and connect a voltmeter in parallel across the conductor.

Using the rheostat, take several pairs of ammeter reading $I$I and voltmeter reading $V$V. Plot $V$V against $I$I; a straight line through the origin verifies $V\propto I$V I, i.e. Ohm's law. The slope gives the resistance.

(b) Compute the ratio $\dfrac{V}{I}$V/I for each pair:

$\dfrac{1.0}{0.25}=4$1.0/0.25=4, $\dfrac{2.0}{0.50}=4$2.0/0.50=4, $\dfrac{3.0}{0.75}=4$3.0/0.75=4, $\dfrac{4.0}{1.00}=4$4.0/1.00=4.

Since $\dfrac{V}{I}=4\,\Omega$V/I=4\, is constant, $V\propto I$V I and the conductor obeys Ohm's law.

The resistance of the conductor is $R=4\,\Omega$R=4\,.

(i) They are connected in parallel so that (a) each appliance gets the same full mains voltage of $230\,\text{V}$230\,V and works independently, and (b) each can be switched on or off separately without affecting the others; if one fails the rest keep working.

(ii) Bulb: $I=\dfrac{V}{R}=\dfrac{230}{460}=0.5\,\text{A}$I=V/R=230/460=0.5\,A.

(iii) Heater: $I=\dfrac{V}{R}=\dfrac{230}{46}=5\,\text{A}$I=V/R=230/46=5\,A.

(iv) The heater dissipates more power. At the same voltage $P=\dfrac{V^2}{R}$P=V^2/R, so the smaller resistance draws more current and dissipates more power: heater $P=\dfrac{230^2}{46}=1150\,\text{W}$P=230^2/46=1150\,W versus bulb $P=\dfrac{230^2}{460}=115\,\text{W}$P=230^2/460=115\,W.