Question 1

Q. 3.5.

At room temperature ( $27.0^{\circ} \mathrm{C}$ ) the resistance of a heating element is $100 \Omega$. What is the temperature of the element if the resistance is found to be $117 \Omega$, given that the temperature coefficient of the material of the resistor is $1.70 	imes 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$

Accepted Answer

Room temperature, $\mathrm{T}=27^{\circ} \mathrm{C}$

Resistance of the heating element at $\mathrm{T}, \mathrm{R}=100 \Omega$

Let $T_{1}$ is the increased temperature of the filament.

Resistance of the heating element at $T_{1}, R_{1}=117 \Omega$

Temperature co-efficient of the material of the filament,

$\alpha=1.70 	imes 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$

$\alpha$ is given by the relation,

$\alpha=\frac{R_{1}-R}{R\left(T_{1}-Tight)}$

$T_{1}-T=\frac{R_{1}-R}{R \alpha}$

$T_{1}-27=\frac{117-100}{100\left(1.7 	imes 10^{-4}ight)}$

$T_{1}-27=1000$

$T_{1}=1027^{\circ} \mathrm{C}$

Therefore, at $1027^{\circ} \mathrm{C}$, the resistance of the element is $117 \Omega$.

Question 2

Q. 3.6.

A negligibly small current is passed through a wire of length 15 m and uniform crosssection $6.0 	imes 10^{-7} \mathrm{~m}^{2}$, and its resistance is measured to be $5.0 \Omega$. What is the resistivity of the material at the temperature of the experiment?

Accepted Answer

Length of the wire, $\mathrm{I}=15 \mathrm{~m}$

Area of cross-section of the wire, $\mathrm{a}=6.0 	imes 10^{-7} \mathrm{~m}^{2}$

Resistance of the material of the wire, R = 5.0 $\Omega$

Resistivity of the material of the wire $=ho$

Resistance is related with the resistivity as

$$

\begin{aligned}

R & =ho \frac{l}{A} \

ho & =\frac{R A}{l} \

& =\frac{5 	imes 6 	imes 10^{-7}}{15}=2 	imes 10^{-7} \Omega \mathrm{~m}

\end{aligned}

$$

Therefore, the resistivity of the material is $2 	imes 10^{-7} \Omega \mathrm{~m}$.

Question 3

Q. 3.7.

A silver wire has a resistance of $2.1 \Omega$ at $27.5^{\circ} \mathrm{C}$, and a resistance of $2.7 \Omega$ at $100^{\circ} \mathrm{C}$.

Determine the temperature coefficient of resistivity of silver.

Accepted Answer

Temperature, $\mathrm{T}_{1}=27.5^{\circ} \mathrm{C}$

Resistance of the silver wire at $T_{1}, R_{1}=2.1 \Omega$

Temperature, $\mathrm{T}_{2}=100^{\circ} \mathrm{C}$

Resistance of the silver wire at $T_{2}, R_{2}=2.7 \Omega$

Temperature coefficient of silver $=a$

It is related with temperature and resistance as

$$

\begin{aligned}

\alpha & =\frac{R_{2}-R_{1}}{R_{1}\left(T_{2}-T_{1}ight)} \

& =\frac{2.7-2.1}{2.1(100-27.5)}=0.0039^{\circ} \mathrm{C}^{-1}

\end{aligned}

$$

Therefore, the temperature coefficient of silver is $0.0039^{\circ} \mathrm{C}^{-1}$.

Question 4

Q. 3.8.

A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A . What is the steady temperature of the heating element if the room temperature is $27.0^{\circ} \mathrm{C}$ ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is $1.70 	imes 10^{-4}{ }^{

Accepted Answer

Supply voltage, $\mathrm{V}=230 \mathrm{~V}$

Initial current drawn, $\mathrm{I}_{1}=3.2 \mathrm{~A}$

Initial resistance $=\mathrm{R}_{1}$, which is given by the relation,

$$

\begin{aligned}

R_{1} & =\frac{V}{I} \

& =\frac{230}{3.2}=71.87 \Omega

\end{aligned}

$$

Steady state value of the current, $\mathrm{I}_{2}=2.8 \mathrm{~A}$

Resistance at the steady state $=\mathrm{R}_{2}$, which is given as $R_{2}=\frac{230}{2.8}=82.14 \Omega$

Temperature co-efficient of nichrome, $\mathrm{a}=1.70 	imes 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$

Initial temperature of nichrome, $\mathrm{T}_{1}=27.0^{\circ} \mathrm{C}$

Study state temperature reached by nichrome $=\mathrm{T}_{2}$

$\mathrm{T}_{2}$ can be obtained by the relation for a ,

$\alpha=\frac{R_{2}-R_{1}}{R_{1}\left(T_{2}-T_{1}ight)}$

$T_{2}-27^{\circ} \mathrm{C}=\frac{82.14-71.87}{71.87 	imes 1.7 	imes 10^{-4}}=840.5$

$T_{2}=840.5+27=867.5^{\circ} \mathrm{C}$

Therefore, the steady temperature of the heating element is $867.5^{\circ} \mathrm{C}$

Question 5

Q. 3.1.

The storage battery of a car has an emf of 12 V . If the internal resistance of the battery is $0.4 \Omega$, what is the maximum current that can be drawn from the battery?

Accepted Answer

Emf of the battery, $\mathrm{E}=12 \mathrm{~V}$

Internal resistance of the battery, $\mathrm{r}=0.4 \Omega$

Maximum current drawn from the battery $=\mathrm{I}$

According to Ohm's law,

$$

E=I r

$$

\begin{aligned}

I & =\frac{E}{r} \

& =\frac{12}{0.4}=30 \mathrm{~A}

\end{aligned}

$$

The maximum current drawn from the given battery is 30 A .

Question 6

Q. 3.2.

A battery of emf 10 V and internal resistance $3 \Omega$ is connected to a resistor. If the current in the circuit is 0.5 A , what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Accepted Answer

Emf of the battery, $\mathrm{E}=10 \mathrm{~V}$

Internal resistance of the battery, $\mathrm{r}=3 \Omega$

Current in the circuit, $\mathrm{I}=0.5 \mathrm{~A}$

Resistance of the resistor $=\mathrm{R}$

The relation for current using Ohm's law is,

$I=\frac{E}{R+r}$

$R+r=\frac{E}{I}$

$$

=\frac{10}{0.5}=20 \Omega

$$

$	herefore R=20-3=17 \Omega$

Terminal voltage of the resistor $=\mathrm{V}$

According to Ohm's law,

$\mathrm{V}=\mathrm{IR}$

$=0.5 	imes 17$

$=8.5 \mathrm{~V}$

Therefore, the resistance of the resistor is $17 \Omega$ and the terminal voltage is 8.5 V .

Question 7

Q. 3.3.

a) Three resistors $1 \Omega, 2 \Omega$, and $3 \Omega$ are combined in series. What is the total resistance of the combination?

b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Accepted Answer

(a) Three resistors of resistances $1 \Omega, 2 \Omega$, and $3 \Omega$ are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances.

Total resistance $=1+2+3=6 \Omega$

(b) Current flowing through the circuit $=\mathrm{I}$

Emf of the battery, $\mathrm{E}=12 \mathrm{~V}$

Total resistance of the circuit, R = $6 \Omega$

The relation for current using Ohm's law is,

$I=\frac{E}{R}$

$$

=\frac{12}{6}=2 \mathrm{~A}

$$

Potential drop across $1 \Omega$ resistor $=\mathrm{V}_{1}$

From Ohm's law, the value of $\mathrm{V}_{1}$ can be obtained as

$\mathrm{V}_{1}=2 	imes 1=2 \mathrm{~V} \ldots$ (i)

Potential drop across $2 \Omega$ resistor $=\mathrm{V}_{2}$

Again, from Ohm's law, the value of $\mathrm{V}_{2}$ can be obtained as

$\mathrm{V}_{2}=2 	imes 2=4 \mathrm{~V} \ldots$ (ii)

Potential drop across $3 \Omega$ resistor $=\mathrm{V}_{3}$

Again, from Ohm's law, the value of $\mathrm{V}_{3}$ can be obtained as

$\mathrm{V}_{3}=2 	imes 3=6 \mathrm{~V} \ldots$ (iii)

Therefore, the potential drop across $1 \Omega, 2 \Omega$, and $3 \Omega$ resistors are $2 \mathrm{~V}, 4 \mathrm{~V}$, and 6 V respectively.

Question 8

Q. 3.4.

(a) Three resistors $2 \Omega, 4 \Omega$ and $5 \Omega$ are combined in parallel. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

Accepted Answer

(a) There are three resistors of resistances,

$\mathrm{R}_{1}=2 \Omega, \mathrm{R}_{2}=4 \Omega$, and $\mathrm{R}_{3}=5 \Omega$

They are connected in parallel. Hence, total resistance (R) of the combination is given by,

$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

$$

=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{10+5+4}{20}=\frac{19}{20}

$$

$	herefore R=\frac{20}{19} \Omega$

Therefore, total resistance of the combination is $\frac{20}{19} \Omega$.

(b) Emf of the battery, $\mathrm{V}=20 \mathrm{~V}$

Current ( $\mathrm{I}_{1}$ ) flowing through resistor $\mathrm{R}_{1}$ is given by,

$$

\begin{aligned}

I_{1} & =\frac{V}{R_{1}} \

& =\frac{20}{2}=10 \mathrm{~A}

\end{aligned}

$$

Current ( $\mathrm{I}_{2}$ ) flowing through resistor $\mathrm{R}_{2}$ is given by,

$$

\begin{aligned}

I_{2} & =\frac{V}{R_{2}} \

& =\frac{20}{4}=5 \mathrm{~A}

\end{aligned}

$$

Current ( $\mathrm{I}_{3}$ ) flowing through resistor $\mathrm{R}_{3}$ is given by,

$$

\begin{aligned}

I_{3} & =\frac{V}{R_{3}} \

& =\frac{20}{5}=4 \mathrm{~A}

\end{aligned}

$$

Total current, $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}=10+5+4=19 \mathrm{~A}$

Therefore, the current through each resister is $10 \mathrm{~A}, 5 \mathrm{~A}$, and 4 A respectively and the total current is 19 A .

3. Current Electricity

Exercises - 1

Exercises - 2