Electricity

3.8 Given

Supply voltage, V = 230 V

Initial current drawn, I= 3.2 A

Final current drawn, $I_1$ = 2.8 A

Room temperature, T = 27.0 °C

Steady temperature, $T_1$ = ?

From Ohm’s law, we get initial resistance, $R$ = $\frac{V}{I}$ = $\frac{230}{3.2}$ Ω = 71.875 Ω

Final resistance, $R_1$ = 

3.7 Let us assume R = 2.1 Ω, T = 27.5 °C, $R_1$ = 2.7 Ω, $T_1$ = 100 $?$ , α = resistivity of silver

We know the relation of α can be given as

α = $\frac{R_1-R}{R(T_1-T)}$ = $\frac{2.7-2.1}{2.1(100-27.5)}$ = 3.94 $\times {10}^{-3}$ ${?}^{-1}$

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

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

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

$\rho =\frac{AR}{l}$ = $\frac{6.0\mathrm{}\times {10}^{-7}\times 5}{15}$ = 2 $\times {10}^{-7}$ 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&nb

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 $R_{1,}$ $I_1$&nbs

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 $\times R$ = 2 $\times 1$ = 2 V

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

Potenti

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

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 p

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