Answer

(b) Current per unit area (taken normal to the current), $I / A$, is called current density and is denoted by $j$. The SI units of the current density are A $/ \mathrm{m}^{2}$. The current density is also directed along $E$ and is also a vector and the relationship is given by

$$ j=s E $$

The $j$ changes due to electric field produced by charges accumulated on the surface of wire.

Note That as the currrent progresses along the wire, the direction of $j$ (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for this.

Answer

(a) The equivalent emf of this combination is given by

$$ \varepsilon_{e q}=\frac{\varepsilon_{2} r_{1}+\varepsilon_{1} r_{2}}{r_{1}+r_{2}} $$

This suggest that the equivalent emf $\varepsilon_{\mathrm{eq}}$ of the two cells is given by

$$ \varepsilon_{1}<\varepsilon_{\mathrm{eq}}<\varepsilon_{2} $$

Thinking Process

Here, the concept of accurate balanced Wheatstone bridge is to be used.

Answer

(c) The percentage error in $R$ can be minimised by adjusting the balance point near the middle of the bridge, i.e., when $I_{1}$ is close to $50 \mathrm{~cm}$. This requires a suitable choice of $S$.

Since,

$$ \frac{R}{S}=\frac{R l_{1}}{R\left(100-l_{1}\right)}=\frac{l_{1}}{100-l_{1}} $$

Since here, $R: S:: 2.9: 97.1$ imply that the $S$ is nearly 33 times to that of $R$. In orded to make this ratio $1: 1$, it is necessary to reduce the value of $S$ nearly $\frac{1}{33}$ times i.e., nearly $3 \Omega$.

Thinking Process

The potential drop across wires of potentiometer should be greater than emfs of primary cells.

Answer

(b) In a potentiometer experiment, the emf of a cell can be measured, if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. Here, values of emfs of two cells are given as $5 \mathrm{~V}$ and $10 \mathrm{~V}$, therefore, the potential drop along the potentiometer wire must be more than $10 \mathrm{~V}$.

Thinking Process

The resistance of wire depends on its geometry $l$ (length of the rod). Here, the metallic rod behaves as a wire.

Answer

(a) The resistance of wire is given by

$$ R=\rho \frac{l}{A} $$

For greater value of $R, l$ must be higher and $A$ should be lower and it is possible only when the battery is connected across $1 \mathrm{~cm} \times \frac{1}{2} \mathrm{~cm}$ (area of cross-section $A$ ).

Answer

(a) The relationship between current and drift speed is given by

$$ I=n e A v_{d} $$

Here, $I$ is the current and $v_{d}$ is the drift velocity.

So,

$$ I \propto v_{d} $$

Thus, only drift velocity determines the current in a conductor.

Answer

$(b, d)$

Kirchhoff’s junction rule is also known as Kirchhoff’s current law which states that the algebraic sum of the currents flowing towards any point in an electric network is zero. i.e., charges are conserved in an electric network.

So, Kirchhoff’s junction rule is the reflection of conservation of charge

Answer

$(a, d)$

Here, the potential drop is taking place across $A B$ and $r$. Since the equivalent resistance of parallel combination of $R$ and $R^{\prime}$ is always less than $R$, therefore $I \geq \frac{V}{r+R}$ always.

Note In parallel combination of resistances, the equivalent resistance is smaller than smallest resistance present in combination.

Answer

$(a, b)$

The resistivity of a metallic conductor is given by,

$$ e=\frac{m}{n e^{2} \tau} $$

where $n$ is number of charge carriers per unit volume which can change with temperature $T$ and $\tau$ is time interval between two successive collisions which decreases with the increase of temperature.

Answer

$(b, c)$

Given, for first student, $R_{2}=10 \Omega, R_{1}=5 \Omega, R_{3}=5 \Omega$

For second student, $R_{1}=500 \Omega, R_{3}=5 \Omega$

Now, according to Wheatstone bridge rule,

$$ \frac{R_{2}}{R}=\frac{R_{1}}{R_{3}} \Rightarrow R=R_{3} \times \frac{R_{2}}{R_{1}} $$

Now putting all the values in E(i.), we get $R=10 \Omega$ for both students. Thus, we can analyse that the Wheatstone bridge is most sensitive and accurate if resistances are of same value.

Thus, the errors of measurement of the two students depend on the accuracy and sensitivity of the bridge, which inturn depends on the accuracy with which $R_{2}$ and $R_{1}$ can be measured.

When $R_{2}$ and $R_{1}$ are larger, the currents through the arms of bridge is very weak. This can make the determination of null point accurately more difficult.

Answer

$(a, c)$

At neutral point, potential at $B$ and neutral point are same. When jockey is placed at to the right of $D$, the potential drop across $A D$ is more than potential drop across $A B$, which brings the potential of point $D$ less than that of $B$, hence current flows from $B$ to $D$.

Answer

When an electron approaches a junction, in addition to the uniform electric field $\mathbf{E}$ facing it normally. It keep the drift velocity fixed as drift velocity depend on $E$ by the relation drift velocity

$$ v_{d}=\frac{e E \tau}{m} $$

This result into accumulation of charges on the surface of wires at the junction. These produce additional electric field. These fields change the direction of momentum.

Thus, the motion of a charge across junction is not momentum conserving.

Thinking Process

The higher drift velocities of electrons make collisions more frequent which in turn decreases the time interval between two successive collision.

Answer

Relaxation time is inversely proportional to the velocities of electrons and ions. The applied electric field produces the insignificant change in velocities of electrons at the order of $1 \mathrm{~mm} / \mathrm{s}$, whereas the change in temperature $(T)$, affects velocities at the order of $10^{2} \mathrm{~m} / \mathrm{s}$. This decreases the relaxation time considerably in metals and consequently resistivity of metal or conductor increases as .

$$ \rho=\frac{1}{\sigma}=\frac{m}{n e^{2} \tau} $$

Answer

The advantage of null point method in a Wheatstone bridge is that the resistance of galvanometer does not affect the balance point, there is no need to determine current in resistances and the internal resistance of a galvanometer.

It is easy and convenient method for observer.

The $R_{\text {unknown }}$ can be calculated applying Kirchhoff’s rules to the circuit. We would need additional accurate measurement of all the currents in resistances and galvanometer and internal resistance of the galvanometer.

Note The necessary and sufficient condition for balanced Wheatstone bridge is

$$ \frac{P}{Q}=\frac{R}{S} $$

where $P$ and $Q$ are ratio arms and $R$ is known resistance and $S$ is unknown resistance.

Answer

In potentiometer, the thick metallic strips are used as they have negligible resistance and need not to be counted in the length $l_{1}$ of the null point of potentiometer. It is for the convenience of experimenter as he measures only their lengths along the straight wires each of lengths $1 \mathrm{~m}$.

This measurements is done with the help of centimetre scale or metre scale with accuracy.

Thinking Process

The availability, conductivity and the cost of the metal are main criterion for the selection of metal for wiring in home.

Answer

The Cu wires or Al wires are used for wiring in the home.

The main considerations are involved in this process are cost of metal, and good conductivity of metal.

Answer

Alloys have small value of temperature coefficient of resistance with less temperature sensitivity.

This keeps the resistance of the wire almost constant even in small temperature change. The alloy also has high resistivity and hence high resistance, because for given length and cross-section area of conductor. ( $L$ and $A$ are constant)

$$ R \propto \rho $$

Answer

The power consumption in transmission lines is given by $P=i^{2} R_{c}$, where $R_{c}$ is the resistance of transmission lines. The power is given by

$$ P=V I $$

The given power can be transmitted in two ways namely (i) at low voltage and high current or (ii) high voltage and low current. In power transmission at low voltage and high current more power is wasted as $P \propto i^{2}$ whereas power transmission at high voltage and low current facilitates the power transmission with minimal power wastage.

The power wastage can be reduced by transmitting power at high voltage.

Answer

With the increase of $R$, the current in main circuit decreases which in turn, decreases the potential difference across $A B$ and hence potential gradient $(k)$ across $A B$ decreases.

Since, at neutral point, for given emf of cell, $I$ increases as potential gradient (k) across $A B$ has decreased because

$$ E=k I $$

Thus, with the increase of $I$, the balance point neutral point will shift towards $B$.

Answer

(i) The deflection in galvanometer is one sided and the deflection decreased, while moving from one end ’ $A$ ’ of the wire to the end ’ $B$ ‘, thus imply that current in auxiliary circuit (lower circuit containing primary cell) decreases, while potential difference across $A$ and jockey increases.

This is possible only when positive terminal of the cell $E_{1}$, is connected at $X$ and $E_{1}>E$.

(ii) The deflection in galvanometer is one sided and the deflection increased, while moving from one end $A$ of the wire to the end $B$, this imply that current in auxiliary circuit (lower circuit containing primary cell) increases, while potential difference across $A$ and jockey increases.

This is possible only when negative terminal of the cell $E_{1}$, is connected at $X$.

Thinking Process

When the cell of emf $E$ and internal resistance $r$ is connected across an external resistance $R$, the relationship between the voltage across $R$ is given by

$$ V=\frac{E}{1+\frac{r}{R}} $$

With the increase of $R, V$ approaches closer to $E$ and when $E$ is infinite, $V$ reduces to 0.

Answer

The graphical relationship between voltage across $R$ and the resistance $R$ is given below

Thinking Process

Here, in series combination of resistors, the equivalent resistance of series combination is in series with the internal resistance $R$ of battery resistors whereas in parallel combination of resistors, the equivalent resistance of parallel combination is in series with the internal resistance of battery.

Answer

In series combination of resistors, current $I$ is given by $I=\frac{E}{R+n R^{\prime}}$

whereas in parallel combination current $10 I$ is given by

$$ \frac{E}{R+\frac{R}{n}}=10 I $$

Now, according to problem,

$$ \frac{1+n}{1+\frac{1}{n}} \Rightarrow 10=\frac{1+n}{n+1} n \Rightarrow n=10 $$

Answer

When all resistances are connected in parallel, the resultant resistance $R_{p}$ is given by

$$ \frac{1}{R_{p}}=\frac{1}{R_{1}}+\ldots \ldots \ldots+\frac{1}{R_{n}} $$

On multiplying both sides by $R_{\min }$ we have

$$ \frac{R_{\min }}{R_{p}}=\frac{R_{\min }}{R_{1}}+\frac{R_{\min }}{R_{2}}+\ldots . .+\frac{R_{\min }}{R_{n}} $$

Here, in RHS, there exist one term $\frac{R_{\text {min }}}{R_{\text {min }}}=1$ and other terms are positive, so we have

$$ \frac{R_{\min }}{R_{p}}=\frac{R_{\min }}{R_{1}}+\frac{R_{\min }}{R_{2}}+\ldots .+\frac{R_{\min }}{R_{n}}>1 $$

This shows that the resultant resistance $R_{p}<R_{\text {min }}$.

Thus, in parallel combination, the equivalent resistance of resistors is less than the minimum resistance available in combination of resistors. Now, in series combination, the equivalent resistant is given by

$$ R_{s}=R_{1}+\ldots \ldots+R_{n} $$

Here, in RHS, there exist one term having resistance $R_{\max }$.

So, we have

or

$$ \begin{aligned} & R_{s}=R_{1}+\ldots+R_{\max }+. .+\ldots+R_{n} \newline \ & R_{s}=R_{1}+\ldots+R_{\max } \ldots+R_{n}=R_{\max }+\ldots\left(R_{1}+\ldots+\right) R_{n} \newline \ & R_{s} \geq R_{\max } \newline \ & R_{s}=R_{\max }\left(R_{1}+\ldots+R_{n}\right) \end{aligned} $$

$$ \text { or } $$

Thus, in series combination, the equivalent resistance of resistors is greater than the maximum resistance available in combination of resistors. Physical interpretation

(a)

(b)

In Fig. (b), $R_{\text {min }}$ provides an equivalent route as in Fig. (a) for current. But in addition there are $(n-1)$ routes by the remaining $(n-1)$ resistors. Current in Fig. (b) is greater than current in Fig. (a). Effective resistance in Fig. (b) $<R_{\text {min }}$. Second circuit evidently affords a greater resistance.

(c)

(d)

In Fig. (d), $R_{\max }$ provides an equivalent route as in Fig. (c) for current. Current in Fig. (d) < current in Fig. (c). Effective resistance in Fig. (d) $>R_{\max }$. Second circuit evidently affords a greater resistance.

Thinking Process

Here, after finding the electric current flow in the circuit by using Kirchhoffs law or Ohm’s law, the potential difference across $A B$ can be obtained.

Answer

Applying Ohm’s law.

Effective resistance $=2 \Omega+8 \Omega=10 \Omega$ and effective emf of two cells $=6-4=2 \mathrm{~V}$, so the electric current is given by

$$ I=\frac{6-4}{2+8}=0.2 \mathrm{~A} $$

along anti-clockwise direction, since $E_{1}>E_{2}$.

The direction of flow of current is always from high potential to low potential. Therefore $V_{B}>V_{A}$.

$\Rightarrow$

Therefore,

$$ \begin{aligned} V_{B}-4 V-(0.2) \times 8 & =V_{A} \ V_{B}-V_{A} & =3.6 V \end{aligned} $$

Thinking Process

Here, after finding the electric current flow in the circuit by using Kirchhoff ’s law or Ohm’s law, the potential difference across first cell can be obtained.

Answer

Applying Ohm’s law,

Effective resistance $=R+r_{1}+r_{2}$ and effective emf of two cells $=E+E=2 E$, so the electric current is given by

$$ I=\frac{E+E}{R+r_{1}+r_{2}} $$

The potential difference across the terminals of the first cell and putting it equal to zero.

or

$$ \begin{aligned} & V_{1}=E-I r_{1}=E-\frac{2 E}{r_{1}+r_{2}+R} r_{1}=0 \newline \newline \ & E=\frac{2 E r_{1}}{r_{1}+r_{2}+R} \Rightarrow 1=\frac{2 r_{1}}{r_{1}+r_{2}+R} \newline \newline \ & r_{1}+r_{2}+R=2 r_{1} \Rightarrow R=r_{1}-r_{2} \end{aligned} $$

This is the required relation.

Answer

The resistance of first conductor

$$ R_{A}=\frac{\rho l}{\pi\left(10^{-3} \times 0.5\right)^{2}} $$

The resistance of second conductor,

$$ R_{B}=\frac{\rho l}{\pi\left[\left(10^{-3}\right)^{2}-\left(0.5 \times 10^{-3}\right)^{2}\right]} $$

Now, the ratio of two resistors is given by

$$ \frac{R_{A}}{R_{B}}=\frac{\left(10^{-3}\right)^{2}-\left(0.5 \times 10^{-3}\right)^{2}}{\left(0.5 \times 10^{-3}\right)^{2}}=3: 1 $$

Thinking Process

The electric current in two cases is obtained and then shown equal to each other

Answer

Let the effective internal resistance of the battery is $R_{\text {eff }}$, the effective external resistance $R$ and the effective voltage of the battery is $V_{\text {eff }}$.

Applying Ohm’s law,

Then current through $R$ is given by

$$ I=\frac{V_{\text {eff }}}{R_{\text {eff }}+R} $$

If all the resistances and the effective voltage are increased $n$-times, then we have

and

$$ \begin{aligned} V_{\text {eff }}^{\text {new }} & =n V_{\text {eff }}, R_{\text {eff }}^{\text {new }}=n R_{\text {eff }} \ R^{\text {new }} & =n R \end{aligned} $$

Then, the new current is given by

$$ I^{\prime}=\frac{n V_{\text {eff }}}{n R_{\text {eff }}+n R}=\frac{n\left(V_{\text {eff }}\right)}{n\left(R_{\text {eff }}+R\right)}=\frac{\left(V_{\text {eff }}\right)}{\left(R_{\text {eff }}+R\right)}=I $$

Thus, current remains the same.

Answer

Applying Kirchhoff’s junction rule, $\quad I_{1}=I+I_{2}$

Applying Kirchhoff’s II law/loop rule applied in outer loop containing $10 \mathrm{~V}$ cell and resistance $R$, we have

$$ 10=I R+10 I_{1} $$

Applying Kirchhoff II law / loop rule applied in outer loop containing 2V cell and resistance $R$, we have

or

$$ \begin{aligned} 2 & =5 I_{2}-R I=5\left(I_{1}-I\right)-R I \ 4 & =10 I_{1}-10 I-2 R I \end{aligned} $$

Solving Eqs. (i) and (ii), gives $\Rightarrow$

$$ \begin{aligned} & 6=3 R I+10 I \ & 2=I R+\frac{10}{3} \end{aligned} $$

Also, the external resistance is $R$. The Ohm’s law states that

$$ V=I\left(R+R_{\text {eff }}\right) $$

On comparing, we have $V=2 V$ and effective internal resistance

$$ \left(R_{\text {eff }}\right)=\frac{10}{3} \Omega $$

Since, the effective internal resistance $\left(R_{\text {eff }}\right)$ of two cells is $\frac{10}{3} \Omega$, being the parallel combination of $5 \Omega$ and $10 \Omega$. The equivalent circuit is given below

Thinking Process

The power consumption in a current carrying resistor is given by $P=I^{2} R$

Answer

Power consumption in a day i.e., in $5=10$ units

Or power consumption per hour $=$ 2units

Or power consumption $=2$ units $=2 \mathrm{~kW}=2000 \mathrm{~J} / \mathrm{s}$

Also, we know that power consumption in resistor,

$\Rightarrow \quad 2000 \mathrm{~W}=220 \mathrm{~V} \times l$ or $l \approx 9 \mathrm{~A}$

Now, the resistance of wire is given by $R=\rho \frac{l}{A}$

where, $A$ is cross-sectional area of conductor.

Power consumption in first current carrying wire is given by

$$ \begin{aligned} P & =l^{2} R \ \rho \frac{l}{A} l^{2} & =1.7 \times 10^{-8} \times \frac{10}{\pi \times 10^{-6}} \times 81 \mathrm{~J} / \mathrm{s} \approx 4 \mathrm{~J} / \mathrm{s} \end{aligned} $$

The fractional loss due to the joule heating in first wire $=\frac{4}{2000} \times 100=0.2 %$

Power loss in Al wire $=4 \frac{\rho_{\mathrm{Al}}}{\rho_{\mathrm{Cu}}}=1.6 \times 4=6.4 \mathrm{~J} / \mathrm{s}$

The fractional loss due to the joule heating in second wire $=\frac{6.4}{2000} \times 100=0.32 %$

Answer

Let $R^{\prime}$ be the resistance of the potentiometer wire.

Effective resistance of potentiometer and variable resistor $(R=50 \Omega)$ is given by $=50 \Omega+R^{\prime}$

Effective voltage applied across potentiometer $=10 \mathrm{~V}$.

The current through the main circuit,

$$ I=\frac{V}{50 \Omega+R}=\frac{10}{50 \Omega+R} $$

Potential difference across wire of potentiometer,

$$ I R^{\prime}=\frac{10 R^{\prime}}{50 \Omega+R} $$

Since with $50 \Omega$ resistor, null point is not obtained it’s possible only when

$$ \Rightarrow \quad \begin{array}{cc} \frac{10 \times R^{\prime}}{50+R} & <8 \ \Rightarrow & 10 R^{\prime}<400+8 R^{\prime} \ 2 R^{\prime}<400 \text { or } R^{\prime}<200 \Omega . \end{array} $$

Similarly with $10 \Omega$ resistor, null point is obtained its possible only when

$$ \begin{array}{rlrl} \frac{10 \times R^{\prime}}{10+R^{\prime}} >8 \newline\newline\ \Rightarrow 2 R^{\prime} >80 \newline \newline\ \Rightarrow \frac{R^{\prime}}{}>40 \newline \newline\ \frac{10 \times \frac{3}{4} R^{\prime}}{10+R^{\prime}} <8 \newline \newline\ \Rightarrow 7.5 R^{\prime} <80+8 R^{\prime} \newline \newline\ R^{\prime} >160 \newline \newline\ 160 <R^{\prime}<200 . \end{array} $$

Any $R^{\prime}$ between $160 \Omega$ and $200 \Omega$ will achieve.

Since, the null point on the last (4th) segment of the potentiometer, therefore potential drop across $400 \mathrm{~cm}$ of wire $>8 \mathrm{~V}$.

This imply that potential gradient

or

$k \times 400 \mathrm{~cm}>8 \mathrm{~V}$

$k \times 4 m>8 V$

$k>2 \mathrm{~V} / \mathrm{m}$

Similarly, potential drop across $300 \mathrm{~cm}$ wire $<8 \mathrm{~V}$.

$$ k \times 300 \mathrm{~cm}<8 \mathrm{~V} $$

or

$$ \begin{array}{r} k \times 3 \mathrm{~m}<8 \mathrm{~V}, \quad k<2 \frac{2}{3} \mathrm{~V} / \mathrm{m} \ 2 \frac{2}{3} \mathrm{~V} / \mathrm{m}>\mathrm{k}>2 \mathrm{~V} / \mathrm{m} \end{array} $$

Thus,

Thinking Process

The current in a conductor and drift velocity of electrons are related as $i=n e A v_{d}$, where $v_{d}$ is drift speed of electrons and $n$ is number density of electrons.

Answer

(a) By Ohm’s law, current $I$ is given by

$$ \begin{aligned} & I=6 \mathrm{~V} / 6 \Omega=1 \mathrm{~A} \ & I=n e t A v_{d} \text { or } v_{d}=\frac{i}{n e A} \end{aligned} $$

But,

On substituting the values

For,

$n=$ number of electron $/$ volume $=10^{29} / \mathrm{m}^{3}$ length of circuit $=10 \mathrm{~cm}$, cross-section $=A=(1 \mathrm{~mm})^{2}$

$$ \begin{aligned} v_{d} & =\frac{1}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}} \ & =\frac{1}{1.6} \times 10^{-4} \mathrm{~m} / \mathrm{s} \end{aligned} $$

Therefore, the energy absorbed in the form of KE is given by

$$ \begin{aligned} \mathrm{KE} & =\frac{1}{2} m_{e} v_{d}^{2} \times n A I \newline \ & =\frac{1}{2} \times 9.1 \times 10^{31} \times \frac{1}{2.56} \times 10^{20} \times 10^{8} \times 10^{6} \times 10^{1} \newline \ & =2 \times 10^{-17} \mathrm{~J} \end{aligned} $$

(b) Power loss is given by $P=I^{2} R=6 \times 1^{2}=6 \mathrm{~W}=6 \mathrm{~J} / \mathrm{s}$ Since, $P=\frac{E}{t}$ Therefore, $E=P \times t$ or

$$ t=\frac{E}{P}=\frac{2 \times 10^{-17}}{6} \approx 10^{-17} \mathrm{~s} $$