Electricity

3.6 Given, length of the wire, l = 15 m

Uniform cross section, A = 6.0 $*{10}^{-7}$ $m^2$

Resistance measured, R = 5.0 Ω

From the relation of

R = $\rho \frac{l}{A}$ , where $\rho$ is the resistivity of the material, we get

$\rho =\frac{AR}{l}$ = $\frac{6.0\mathrm{}*{10}^{-7}*5}{15}$ = 2 $*{10}^{-7}$ Ωm.

3.5 Let T be the room temperature and R be the resistance at room temperature and $T_1$ be the required temperature and $R_1$ be the resistance at that temperature and α be the coefficient of resistor.

We have:

T = 27 $?,$ R = 100 Ω, $T_1$ = ?, $R_1$ = 117 Ω, α = 1.70 $*{10}^{-4}$ ° ${\mathit{C}}^{-1}$

We know the relation of α can be given as

α = $\frac{R_1-R}{R(T_1-T)}$ = $\frac{117-100}{100(T_1-27)}$ = 1.70 $*{10}^{-4}$

or ( $T_1$ - 27) = $\frac{117-100}{100*1.70\mathrm{}*{10}^{-4}}$ = 1000

$T_1$ = 1000 +27 = 1027 $?$$$

3.4 (a) Let $R_1$ = 2 Ω, $R_2$ = 4 Ω, $R_3$ = 5 Ω

If the equivalent resistance is R, then $\frac{1}{R}$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$ + $\frac{1}{R_3}$ = $\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{5}$ = $\frac{10+5+4}{20}$ = $\frac{19}{20}$

R = $\frac{20}{19}$ = 1.05 Ω

(b) The EMF of the battery = 20 V

Current through $R_{1,}$ $I_1$ = $\frac{V}{R_{1,}}$ = $\frac{20}{2}$ = 10 A

Current through $R_{2,}$ $I_2$ = $\frac{V}{R_{2,}}$ = $\frac{20}{4}$ = 5A

Current through $R_{3,}$ $I_3$ = $\frac{V}{R_{3,}}$ = $\frac{20}{5}$ = 4 A

Total current I = $I_1$ + $I_2$ + $I_3$ = 10 + 5 + 4 = 19 A

3.3 (a) The equivalent resistance of the resistor in series is given by

R = 1 + 2 + 3 = 6 Ω

(b) From Ohm's law, I = $\frac{V}{R}$ we get I = $\frac{12}{6}$ = 2 A.

Potential drop across 1 Ω resistor = I $*R$ = 2 $*1$ = 2 V

Potential drop across 2 Ω resistor = I $*R$ = 2 $*2$ = 4 V

Potential drop across 3 Ω resistor = I $*R$ = 2 $*3$ = 6 V

3.2 EMF of the battery = 10 V

Internal resistance of the battery, r = 3 Ω

Current in the circuit, I = 0.5 A

Let the resistance of the resistor be R

According to Ohm's law

I = $\frac{E}{(R+r)}$

R + r = $\frac{E}{I}$ or R = $\frac{E}{I}$ - r = $\frac{10}{0.5}$ - 3 = 17 Ω

Terminal voltage of the battery when the circuit is closed is given by

V = IR = 0.5 $*17=8.5V$

Therefore the resistance is 17 Ω and the terminal voltage is 8.5 V

3.1 EMF of the battery, E = 12 V

Internal resistance of the battery, r = 0.4 Ω

Let the maximum current drawn = I

According to OHM's law, E = Ir

So I = $\frac{E}{r}$ = $\frac{12}{0.4}$ amp = 30 amp

Therefore, the maximum current can be drawn is 30 ampere.

Kirchhoff's laws include the Loop Rule (KVL) and the Junction Rule. These are helpful in analyzing complex electrical circuits. The solutions explain the concepts behind these laws and help them in applying in various example problems involving the EMF sources and multiple resistors. By thoroughly practicing these solutions, the students learn how to choose the junctions and loops strategically for solving problems. This also helps in preparing for exams that include circuit analysis.