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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 $\times {10}^{-3}$ ${?}^{-1}$

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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.

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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 $\times {10}^{-4}$ ° ${\mathit{C}}^{-1}$

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 $\times {10}^{-4}$

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

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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

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

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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 $\times 17=8.5V$

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

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CMS College of Engineering and Technology provides good placements to its BTech graduating students. Check out the table below to know the key highlights of the CMS College of Engineering and Technology placement in 2023-2025:

The institution has not released the course-wise placement packages as of yet. Meanwhile check out the table below to know about the overall packages recorded at CMS College of Engineering and Technology placement in 2023-2025: 

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