Class 12th

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, $R_1$ = $\frac{V}{I_1}$ = $\frac{230}{2.8}$ Ω = 82.143 Ω

From the relation of α = $\frac{R_1-R}{R(T_1-T)}$ , where α is the temperature coefficient of resistance, we get

1.7 $*{10}^{-4}$ = $\frac{82.143-71.875}{71.875(T_1-27)}$

$T_1-27=$ 840.34

$T_1=867.34?$

Therefore the steady temperature of heating element required is $867.34?$$$

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 $*{10}^{-3}$ ${?}^{-1}$

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.