Answer

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

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

Maximum current drawn from the battery $=I$

According to Ohm’s law,

$$ \begin{aligned} E & =I r \\ I & =\frac{E}{r} \\ & =\frac{12}{0.4}=30 \mathrm{~A} \end{aligned} $$

The maximum current drawn from the given battery is $30 \mathrm{~A}$.

Answer

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

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

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

Resistance of the resistor $=R$

The relation for current using Ohm’s law is,

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

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

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

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

Terminal voltage of the resistor $=V$

According to Ohm’s law,

$V=I R$

$=0.5 \times 17$

$=8.5 \mathrm{~V}$

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

$8.5 \mathrm{~V}$.

Answer

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

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

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,

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

$\alpha$ is given by the relation,

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

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

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

$T_{1}-27=1000$

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

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

Answer

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

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

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

Resistivity of the material of the wire $=\rho$

Resistance is related with the resistivity as

$$ \begin{aligned} R & =\rho \frac{l}{A} \\ \rho & =\frac{R A}{l} \\ & =\frac{5 \times 6 \times 10^{-7}}{15}=2 \times 10^{-7} \Omega \mathrm{m} \end{aligned} $$

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

Answer

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

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

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

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

Temperature coefficient of silver $=\alpha$

It is related with temperature and resistance as

$$ \begin{aligned} \alpha & =\frac{R_{2}-R_{1}}{R_{1}\left(T_{2}-T_{1}\right)} \\ & =\frac{2.7-2.1}{2.1(100-27.5)}=0.0039^{\circ} \mathrm{C}^{-1} \end{aligned} $$

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

Answer

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

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

Initial resistance $=R_{1}$, 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, $I_{2}=2.8 \mathrm{~A}$

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

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

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

Study state temperature reached by nichrome $=T_{2}$

$T_{2}$ can be obtained by the relation for $\alpha$,

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

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

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

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

Answer

Emf of the storage battery, $E=8.0 \mathrm{~V}$

Internal resistance of the battery, $r=0.5 \Omega$

DC supply voltage, $V=120 \mathrm{~V}$

Resistance of the resistor, $R=15.5 \Omega$

Effective voltage in the circuit $=V^{1}$

$R$ is connected to the storage battery in series. Hence, it can be written as

$V^{1}=V-E$

$V^{1}=120-8=112 \mathrm{~V}$

Current flowing in the circuit $=I$, which is given by the relation,

$$ \begin{aligned} I & =\frac{V^{1}}{R+r} \\ & =\frac{112}{15.5+5}=\frac{112}{16}=7 \mathrm{~A} \end{aligned} $$

Voltage across resistor $R$ given by the product, $I R=7 \times 15.5=108.5 \mathrm{~V}$

DC supply voltage $=$ Terminal voltage of battery + Voltage drop across $R$

Terminal voltage of battery $=120-108.5=11.5 \mathrm{~V}$

A series resistor in a charging circuit limits the current drawn from the external source. The current will be extremely high in its absence. This is very dangerous.

Answer

Number density of free electrons in a copper conductor, $n=8.5 \times 10^{28} \mathrm{~m}^{-3}$ Length of the copper wire, $l=3.0 \mathrm{~m}$

Area of cross-section of the wire, $A=2.0 \times 10^{-6} \mathrm{~m}^{2}$

Current carried by the wire, $I=3.0 \mathrm{~A}$, which is given by the relation,

$I=n A \mathrm{e} V_{\mathrm{d}}$

Where,

$\mathrm{e}=$ Electric charge $=1.6 \times 10^{-19} \mathrm{C}$

$V_{\mathrm{d}}=$ Drift velocity $=\frac{\text { Length of the wire }(l)}{\text { Time taken to cover } l(t)}$

$I=n A \mathrm{e} \frac{l}{t}$

$t=\frac{n A \mathrm{e} l}{I}$

$=\frac{3 \times 8.5 \times 10^{28} \times 2 \times 10^{-6} \times 1.6 \times 10^{-19}}{3.0}$

$=2.7 \times 10^{4} \mathrm{~s}$

Therefore, the time taken by an electron to drift from one end of the wire to the other is $2.7 \times 10^{4} \mathrm{~s}$.