Current Electricity Class 12 Important Questions: CBSE Physics Chapter 3

Very Short Answer Questions: (1 Marks)

1. If the temperature of a good conductor decreases, how does the relaxation time of electrons in the conductor change? 

Ans: It is known that, 

$\rho =\frac{m}{n{e}^2\tau }$ 

Therefore, if the temperature of a good conductor decreases, collision decreases and thus the relaxation time of electrons increases which in turn decreases the resistivity. 

2. If potential difference $V$ applied across a conductor is increased to $2V$, how will the drift velocity of the electron change?

Ans: It is known that,

$V_d=\frac{eE\tau }{m}=\frac{eV\tau }{lm}$

When the potential difference across a conductor is doubled, the drift velocity of the electron gets doubled.

3. What is the value of current $I$ at $O$ in the adjoining circuit?

Ans: Total current at O is given by,

$i=5+3-2-7+8$ 

$\Rightarrow i=16-9$

$\Rightarrow i=7A$

Therefore, the current $I$ at $O$is $i=7A$.

4. State one condition for maximum current to be drawn from the cell?

Ans: It is known that,

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

To find maximum current, internal resistance must be zero.

5. Resistivities of copper, silver and manganin are $1.7\times {10}^{-8}m$, $1.0\times {10}^{-8}m$ and $44\times {10}^{-8}m$ respectively which of these is the best conductor?

Ans: The resistance is directly proportional to specific resistance (resistivity), when length and area of cross-section is made constant.$R=\frac{\rho l}{A}$

Hence, silver is the best conductor as its specific resistance is less.

6. Draw the graph showing the variation of conductivity with temperature for a metallic conductor? 

Ans: The conductivity for a metallic conductor decreases with the increase in temperature.   

7. If a wire is stretched to double its length, what will be its new resistivity? 

Ans: There will be no change in resistivity because resistivity depends only on the nature of the material. 

8. Name any one material having a small value of temperature coefficient of resistance. Write one use of this material?

Ans: Material having a small value of temperature coefficient of resistance is Nichrome. It is an alloy used for making standard resistance coil. 

9. Two wires $A$ and $B$ are of the same metal and of same length have their areas of cross section in the ratio $2:1$ if the same potential difference is applied across each wire in turn, what will be the ratio of current flowing in $A$ & $B$ ?

Ans: It is known that,

$I=\frac{V}{R}$ and $R<\frac{1}{A}$

When area of cross section is $2:1$ then the ratio of current flowing in $A$ & $B$ will be $1:2$.

Very Short Answer Questions: (2 Marks)

1. Two electric bulbs A and B are marked $$\mathbf{220}\mathbf{V}$$,$40W$ and $220V$$60W$ respectively. Which one has a higher resistance?

Ans: It is known that,

$R=\frac{V^2}{p}$ 

For Bulb $A$, $R_1=\frac{{(220)}^2}{40}=1210\mathrm{\Omega }$ 

For Bulb $B$, $R_2=\frac{{(220)}^2}{60}=806.67\mathrm{\Omega }$ 

Therefore, bulb $A$ has higher resistance.

2. A Carbon resistor has three strips of red colour and a gold strip. What is the value of the resistor? What is its tolerance? 

Ans: It is given that,

$$RRRGold=(22\times {10}^2)\pm 5$$

Therefore, value of the resistor$=2200\mathrm{\Omega }$ and tolerance$=\pm 5$

3. Determine the voltage drop across the resistor $R_1$ in the circuit given below with $$\mathbf{E}=\mathbf{60}\mathbf{V}$$, $R_1=18\mathrm{\Omega }$, $R_2=10\mathrm{\Omega }$, $R_3=5\mathrm{\Omega }$ and $R_4=10\mathrm{\Omega }$.

Ans: In the given circuit,

$R_3\mathrm{\&}{R}_4$ are in series

$R^{\prime }=R_3+R_4$

$$\Rightarrow R^{\prime }=5+10=15\mathrm{\Omega }$$

Then, $R^{\prime }\mathrm{\&}{R}_2$ are in parallel

$\frac{1}{R^{″}}=\frac{1}{R^1}+\frac{1}{R_2}$ 

$\Rightarrow \frac{1}{R^{″}}=\frac{1}{15}+\frac{1}{10}$

$\Rightarrow \frac{1}{R^{″}}=\frac{4+6}{60}=\frac{10}{60}$

$\Rightarrow R^{″}=\frac{60}{10}=6\mathrm{\Omega }$

Then $R_1,R^{″}$ are series

$R_{net}=R^{″}+R_1$ 

$\Rightarrow R_{net}=6+18=24\mathrm{\Omega }$ 

Current: $I=\frac{V}{R}=\frac{60}{24}\text{ampere}$ 

Voltage drop across $R_1=I{R}_1=\frac{60}{24}\times 18$ 

$\Rightarrow V_1=45v\text{olts}$

Therefore, voltage drop across the resistor $R_1$ is $45volts$.

4. Two heated wires of the same dimensions are first connected in series and then it’s parallel to a source of supply. What will be the ratio of heat produced in the two cases? 

Ans: It is known that,

$H=I^2\mathrm{R}t$    $(I=\frac{V}{R})$

$\Rightarrow H=\frac{V^2}{R^2}\times R\times t$ 

$\Rightarrow H=\frac{V^2}{R}t$

$\Rightarrow H\propto \frac{1}{R}$

$$\frac{{\text{H}}_{\text{series}}}{{\text{H}}_{\text{parallel}}}=\frac{{\text{R}}_{\text{parallel}}}{{\text{R}}_{\text{series}}}$$

$$\Rightarrow \frac{\frac{1}{(\frac{1}{R}+\frac{1}{R})}}{R+R}=\frac{R/2}{2R}=\frac{R}{2R\times 2}=\frac{1}{4}$$

Therefore, the ratio of heat produced is$1:4$.

5. V.i graph for a metallic wire at two different temperatures is shown in the figure. Which of these two temperatures is higher and why? 

Ans: From the graph: 

$Slope=\frac{i}{V}$

It is known that, $\frac{i}{V}=\frac{1}{R}$

It means that the smaller the slope, the larger the resistance.

As the resistance increases, temperature also increases.

Temperature of $T_2$ is higher because $T_2$ has a small slope.

6. A set of n-identical resistors, each of resistance $R$ ohm when connected in series have an effective resistance of $X$ ohm and when the resistors are connected in parallel the effective resistance is $Y$ ohm. Find the relation between $R$,$X$ and $Y$? 

Ans: It is given that,

n-resistors connected in series 

$\Rightarrow X=nR$    ……$(1)$

n-resistors connected in parallel

$\Rightarrow Y=\frac{R}{n}$     ……$(2)$

Multiply equations $(1)$ and $(2)$

$\Rightarrow XY=nR\times \frac{R}{n}$ 

$\Rightarrow XY=R^2$

$\Rightarrow R=\sqrt{XY}$

7. Show that the resistance of a conductor is given by $R=\frac{ml}{n{e}^2\tau A}$.

Ans: Consider a conductor of length $l$  and area $A$.

If electric field $E$ is applied across the conductor, 

The drift velocity of electrons $v_d$ is given by:

$v_d=\frac{eE}{m}\tau$

It is known that, $I=neA{v}_d$

$\Rightarrow \text{I}=\mathrm{neA}\left(\frac{eE}{m}\tau \right)$; $E=\frac{V}{l}$

$\Rightarrow \text{I}=\mathrm{neA}\left(\frac{eV}{ml}\tau \right)$

$\Rightarrow \frac{V}{I}=\frac{ml}{n{e}^{2}A\tau }$;$\frac{V}{I}=R$

$\Rightarrow R=\frac{m}{n{e}^2\tau }\left(\frac{l}{A}\right)$

$\Rightarrow \text{R}=\frac{ml}{n{e}^2\tau A}$

Therefore, the resistance of the conductor, $\text{R}=\frac{ml}{n{e}^2\tau A}$

8. Figure shows a piece of pure semiconductor $S$ in series with a variable resistor $R$ and a source of constant voltage $V$. Would you increase and decrease the value of $R$ to keep the reading of ammeter $(A)$ constant, when semiconductor $S$ is heated? Give reasons.

Ans: On increasing the temperature, resistance of the semiconductor decreases. So, to increase the temperature, the semiconductor $S$ is heated.

To maintain constant current in ammeter, total resistance of the circuit should remain unchanged.

Therefore, the value of $R$ has to be increased.

9. Why is constantan or manganin used for making standard resistors?

Ans: Constantan or manganin are alloys of high resistivity and low temperature coefficient of resistance. So, these are used for making standard resistors.

10. What are ohmic and non-ohmic resistors? Give one example of each?

Ans: Ohmic resistors are resistors which obey ohm’s law. Eg: Metals 

Non-ohmic resistors are resistors which do not obey ohm’s law. Eg: semiconductor diode, transistor etc.

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

Ans: It is given that,

Emf of the battery, $$E=12V$$ 

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

It is known that,

$V=E-iR$

To get maximum current from that battery:

$\Rightarrow E-iR=0$

$\Rightarrow E=iR$

$\Rightarrow i=\frac{E}{R}=\frac{12}{0.4}$

$\Rightarrow i=30A$

Therefore, the maximum current drawn from the given battery,$i=30A$.

12. In a potentiometer arrangement, a cell of emf $1.25V$ gives a balance point at $35.0cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $63.0cm$, what is the emf of the second cell?

Ans: It is given that,

Emf of the cell, $E_1=1.25V$  

Balance point of the potentiometer, $l_1=35cm$ 

The cell is replaced by another cell of emf  $E_2$ 

New balance point of the potentiometer,  $l_2=63cm$ 

Balance condition of potentiometer: $\frac{E_1}{E_2}=\frac{l_1}{l_2}$ 

$\Rightarrow E_2=E_1\times \frac{l_2}{l_1}$

$\Rightarrow E_2=1.25\times \frac{63}{35}=2.25\text{V}$

Hence, the emf of the second cell $2.25V$.

13. What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Ans: It can be concluded that the ratio of voltage with current is constant and is equal to $19.7$ i.e., it obeys Ohm’s law. Therefore, manganin is an ohmic conductor.

From Ohm’s law, $\frac{V}{I}=R$ i.e., the ratio of voltage with current is the resistance of the conductor. Thus, the resistance of manganin is $19.7\mathrm{\Omega }$.

Short Answer Questions: (3 Marks)

1. What happens to the resistance of the wire when its length is increased to twice its original length?

Ans: It is known that,

$R=\rho \frac{l}{A}=\rho (\frac{l}{\pi r^2})$ 

It is given that, 

New length of the wire, $l^{\prime }=2l$ 

Let, the new radius be $r^{\prime }$.

As volume of the wire remains the same: $\pi {\text{r}}^2\ell =\pi {(r^{\prime })}^2{\ell }^{\prime }$ 

$\Rightarrow \pi r^{2}l=\pi {(r^{\prime })}^{2}2\ell$

$$\Rightarrow {(r^{\prime })}^2=\frac{r^2}{2}$$

New resistance, $R^{\prime }=\rho \left(\frac{l^{\prime }}{\pi r{{}^{\prime }}^2}\right)$

$\Rightarrow R^{\prime }=\rho \frac{2l}{\pi \left(\frac{r^2}{2}\right)}$

$\Rightarrow R^{\prime }=\rho \left(\frac{4l}{\pi r^2}\right)$

$\Rightarrow R^{\prime }=4\rho \left(\frac{l}{\pi r^2}\right)$

$\Rightarrow R^{\prime }=4R$

Hence, new resistance becomes four times the original resistance.

2. Mark the direction of current in the circuit as per Kirchhoff’s first rule. What is the value of the main current in the shown network? 

Ans: In the given figure,

$R_2$ and $R_3$ are in series

$\Rightarrow R=3+3=6\mathrm{\Omega }$ 

$R$ and $R_1$ are in parallel

$\frac{1}{R_{net}}=\frac{1}{R}+\frac{1}{R_1}$ 

$\Rightarrow \frac{1}{R_{net}}=\frac{1}{6}+\frac{1}{3}=\frac{3+6}{18}$

$\Rightarrow R_{net}=\frac{18}{9}=2\mathrm{\Omega }$

Net current: $I=\frac{V}{R}$

$I=\frac{2}{2}=1A$ 

Therefore, net current is $1A$.

3. 

(a) Why do we prefer a potentiometer to measure the emf of a cell rather than a voltmeter?

Ans: Potentiometer is based on a null method i.e.; it draws no current from the cell. Emf of a cell is equal to the terminal potential difference when no current flows from the cell. Therefore, a potentiometer is preferred to measure the emf of a cell rather than a voltmeter.

(b) With suitable circuit diagram, show how emfs of 2 cells can be compared using a potentiometer?

Ans: Potentiometer works on the principle that when a constant current flows through the wire of uniform area of cross-section then

Condition: 

Close the switch 1 and 3 such that $E_1$comes in the circuit.

Potential Difference across AJ is $V_{AJ}<I_1$ 

It is known that, no current flows between $E_1$  and  $V_{AJ}$ 

$\Rightarrow V_{AJ}=E_1$

$E_1<l_1$

$\Rightarrow E_1=k{l}_1$    ……$(1)$

Condition: 

Close the switch 2 and 3 such that $E_2$ comes in the circuit and balance point is obtained of $J_1$ .

Since no current flows between $A{J}_1$ and $E_2$

$\Rightarrow V_{A{J}_1}=E_2$

$E_2<l_2$

$\Rightarrow E_2=k{l}_2$  ……$(2)$

$\Rightarrow V_{AJ}=E_2=k{l}_2$  

Dividing equations $(1)$ and $(2)$

$\Rightarrow \frac{E_1}{E_2}=\frac{k{l}_1}{k{l}_2}$ 

$\Rightarrow \frac{E_1}{E_2}=\frac{l_1}{l_2}$

Therefore, $\frac{E_1}{E_2}=\frac{l_1}{l_2}$.

4. Potential difference $V$ is applied across the ends of copper wire of length $l$ and diameter $D$. What is the effect on drift velocity of electrons if

(a) $$V$$ is doubled 

Ans: It is known that,

$V_d=\frac{I}{neA}=\frac{V}{R(neA)}$

$\Rightarrow V_d=\frac{V}{\left(\rho \frac{\ell }{A}\right)(neA)}=\frac{V}{ne\rho \ell }$

Therefore, when $V$ is doubled, drift velocity gets doubled.

(b) $\ell$ is doubled

Ans:  It is known that,

$V_d=\frac{I}{neA}=\frac{V}{R(neA)}$

$\Rightarrow V_d=\frac{V}{\left(\rho \frac{\ell }{A}\right)(neA)}=\frac{V}{ne\rho \ell }$

Therefore, when $l$ is doubled, drift velocity gets halved.

(c) D is doubled 

Ans: It is known that,

$V_d=\frac{I}{neA}=\frac{V}{R(neA)}$

$\Rightarrow V_d=\frac{V}{\left(\rho \frac{\ell }{A}\right)(neA)}=\frac{V}{ne\rho \ell }$

Therefore, as $V$ is independent of $D$, drift velocity remains unchanged.

5. What is drift velocity? Derive expression for drift velocity of electrons in a good conductor in terms of relaxation time of electrons? 

Ans: Drift velocity is defined as the average velocity with which free electrons get drifted in a direction opposite to that of an electric field.

Let, $m$ be the mass of the electron and $e$ be the charge of electrons.

When electric field $E$ is applied, acceleration acquired by the electron is $a=\frac{eE}{m}$.

From, first equation of motion: $v=u+at$ 

Average initial velocity, $u=OV=v_d$

Relaxation time, $t=\tau$

$\Rightarrow v_d=a\tau$ 

$\Rightarrow v_d=\frac{eE\tau }{m}$

Where,

$e$ is the change on electron 

$E$ is the electric field intensity 

$\tau$  is the relaxation time 

$m$ is the mass of electron

Therefore, the expression for drift velocity of electrons in a good conductor in terms of relaxation time of electrons is $v_d=\frac{eE\tau }{m}$.

6. The potentiometer circuit shown, the balance (null) point is at $X$. State with reason, where the balance point will be shifted when 

(a) Resistance R is increased, keeping all parameters unchanged. 

Ans: If resistance $R$ is increased, then the current through potentiometer wire $AB$ will decrease. Therefore, the potential difference across $A$ will decrease and balance point shifts towards $B$. 

(b) Resistance S is increased, keeping R constant.

Ans: If resistance $S$ is increased, then the terminal potential difference of the battery will decrease. Therefore, the balance point will be obtained at a smaller length and thus shifts towards $A$. 

(c) Cell P is replaced by another cell whose emf is lower than that of cell Q.

Ans: If cell $P$ is replaced by another cell whose emf is lower than that of cell Q, then the potential difference across $AB$ will be less than that of emf Q. Therefore, the balance point will not be obtained.

7.

(a) Using the principle of Wheatstone bridge describes the method to determine the specific resistance of a wire in the laboratory. Draw the circuit diagram and write the formula used? 

Ans: Firstly, close the Key $(k)$ and move the jockey along the wire till a certain point $B$ is reached where the galvanometer shows no deflection. Therefore, the bridge is said to be balanced. 

If $Rcm$is the resistance per cm length of the wire then.

$\frac{R}{X}=\frac{lRcm}{(100-l)Rcm}$ 

$X=\frac{R(100-l)}{l}$

It is known that,

$P=\frac{XA}{l^{\prime }}$, $l^{\prime }$is the length if the wire

$P=\frac{R(100-l)A}{l(l^{\prime })}$ 

Therefore, the point is located at, $P=\frac{R(100-l)A}{l(l^{\prime })}$.

(b) In a Wheatstone bridge experiment, a student by mistake connects key $(k)$ in place of galvanometer and galvanometer $(G)$ in place of Key $(K)$. What will be the change in the deflection of the bridge? 

Ans: If the bridge is balanced, then there will be no current in the key. Therefore, constant current flows through the galvanometer and thus there will be no change in deflection of the bridge.

8. Two primary cells of emf $E_1$ and $E_2$  are connected to the potentiometer wire $AB$ as shown in the figure if the balancing length for the two combinations of the cells are $250cm$ and $400cm$. Find the ratio of $E_1$ and $E_2$ .

Ans: It is known that,$E=kl$

$E_1-E_2=K\times 250$   ……$(1)$

$E_1+E_2=K\times 400$   ……$(2)$

Adding equations $(1)$ and $(2)$

$2E_1=250K+400K$

$$2E_1=250K+400$$ 

$2E_1=650K$ 

$E_1=\frac{650}{2}K$

$E_1=325K$

Substituting $E_1=325K$ in equation$(1)$

$\text{ 325K}-E_2=K\times 250$

$E_2=325K-250K$

$E_2=75K$

Divide $E_1$ and $E_2$

$$\Rightarrow \frac{E_1}{E_2}=\frac{325K}{75K}$$

$$\Rightarrow \frac{E_1}{E_2}=4.33$$

The ratio of $E_1$ and $E_2$ is $4.33$.

9. Explain with the help of a circuit diagram, how the value of an unknown resistance can be determined using a Wheatstone bridge? 

Ans: In this case $$P,Q,R$$ are known resistance and $X$ is an unknown resistance.

Applying Kirchhoff’s law for closed path $$ABDA$$. 

$I_{1}P+I_{3}G-I_{2}R=0$    ……$(1)$

For closed path $$BCDB$$

$(I_1-I_3)Q-(I_2+I_3)X-I_{3}G=0$   …… $(2)$

The bridge is said to be balanced when no current flows through the galvanometer. 

$\Rightarrow Ig=0$ $(I_g=I_3)$

Equation $(1)\Rightarrow I_{1}P=I_{2}R$

$$\Rightarrow \frac{I_1}{I_2}=\frac{R}{P}$$      ……$(3)$

Equation $(2)\Rightarrow I_{1}Q=I_{2}X$

$$\Rightarrow \frac{I_1}{I_2}=\frac{X}{Q}$$      ……$(4)$

Equating equations $(3)$and $(4)$

$\frac{R}{P}=\frac{X}{Q}\Rightarrow X=\frac{RQ}{P}$ 

10. Find the current drawn from a cell of emf  $1V$ and internal resistance $$\frac{2}{3}\mathrm{\Omega }$$ connected to the network shown in the figure. $E=1V$ , $r=\frac{2}{3}\mathrm{\Omega }$.

Ans: The circuit can be redrawn as follows:

$\Rightarrow \frac{I}{R_1}=\frac{1}{1}+\frac{1}{1}$ 

$\Rightarrow \frac{1}{R_1}=2$

$\Rightarrow R_1=\frac{1}{2}$

Similarly,

$\Rightarrow R_2=\frac{1}{2}$

$R=R_1+R_2$

$\Rightarrow R=\frac{1}{2}+\frac{1}{2}$

$\Rightarrow R=1\mathrm{\Omega }$

Here, $$1\mathrm{\Omega },1\mathrm{\Omega }$$ and $$1\mathrm{\Omega }$$ are in parallel.

$\Rightarrow \frac{1}{R_{net}}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}$

$\Rightarrow \frac{1}{Rnet}=\frac{3}{1}$

$\Rightarrow \text{ Rnet }=\frac{3}{1}\mathrm{\Omega }$

$\text{I}=\frac{E}{R+r}=\frac{1}{\frac{3}{1}+\frac{3}{3}}=\frac{3}{3}=1A$

$\Rightarrow I=1A$

Current drawn from a cell of emf  $1V$ is $I=1A$.

11.

(a) State and explain Kirchhoff’s law? 

Ans: Kirchhoff’s first law: It states that the algebraic sum of currents at a junction in an electrical circuit is always zero. 

$\Rightarrow i_1-i_2+i_3-i_4-i_5-i_6=0$ 

Kirchhoff’s second law: It states that in any closed part of an electrical circuit, the algebraic sum of emf is equal to the algebraic sum of the products of resistances and current flowing through them. Eg: For closed path $$ABCA$$

$R_1{I}_1-R_3{I}_3+R_2{I}_2-E_1+E_2=0$ 

$E_1-E_2=R_1{I}_1-R_3{I}_3+R_2{I}_2$

(b) In the network shown, find the values of current $I_1,I_2,I_3$.

Ans: Applying Kirchhoff’s law at point D

$I_1=I_2+I_3$      ……$(1)$

For closed path $$ABDA$$

$2I_1+1-2+I_1+3I_2=0$ 

$3I_1+3I_2-1=0$

$3I_1+3I_2=1$       ……$(2)$

For closed path $$DBCD$$

$3I_2-1-I_3-3I_3+3=0$

$3I_2-4I_3+2=0$

$4I_3-3I_2-2=0$

$4I_3-3I_2=2$     ……$(3)$

Substituting $(1)$ in $(2)$

$(2)\Rightarrow 3(I_2+I_3)+3I_2=1$

$(2)\Rightarrow 6I_2+3I_3=1$

$(3)\Rightarrow 4I_3-3I_2=2$

$2\times (3)\Rightarrow 8I_3-6I_2=4$       ……$(4)$

$(3)+(4)=11I_3=5$

$\Rightarrow I_3=\frac{5}{11}A$

Substituting $I_3$ in equation $(3)$

$\Rightarrow 4(\frac{5}{11})-3I_2=2$

$\Rightarrow \frac{20}{11}-3I_2=2$

$\Rightarrow \frac{20}{11}-2=3I_2$

$\Rightarrow \frac{-2}{11}=3I_2$

$\Rightarrow I_2=\frac{-2}{33}A$

Substituting $I_2$,$I_3$ in equation $(1)$

$I_1=\frac{-2}{33}+\frac{5}{11}$

$\Rightarrow I_1=\frac{13}{33}A$

Currents are $I_1=\frac{13}{33}A$, $I_2=\frac{-2}{33}A$ and $I_3=\frac{5}{11}A$.

12. The variation of resistance of a metallic conductor with temperature is given in figure.

(a) Calculate the temperature coefficient of resistance from the graph.

Ans: It is known that,

Temperature coefficient of Resistance:$\alpha =\frac{R-R_0}{R_0\theta }$ 

where $R$ is the resistance of the conductor and $\theta$ is the temperature corresponding to point A. 

(b) State why the resistance of the conductor increases with the rise in temperature. 

Ans: It is known that, 

$R=\rho \frac{l}{A}=\frac{m}{n{e}^2\tau }(\frac{t}{A})$

where $\rho$is the resistivity.

When temperature increases, the number of collisions increases, average relaxation time decreases, therefore resistance increases. 

13. A circle ring having negligible resistance is used to connect four resistors of resistances $$\mathbf{6}\mathbf{R},\mathbf{6}\mathbf{R},\mathbf{6}\mathbf{R}$$ and $R$ as shown in the figure. Find the equivalent resistance between points A & B.

Ans: The given circuit can be redrawn as follows:

$$6R,6R$$ and $6R$ are in parallel.

$\Rightarrow \frac{1}{R_s}=\frac{1}{6R}+\frac{1}{6R}+\frac{1}{6R}$ 

$\Rightarrow \frac{1}{R_s}=\frac{3}{6R}$

$\Rightarrow R_S=\frac{6R}{3}=2R$ 

$R_S=2R$ and $R$ are in series.

$R_{net}=2R+R$ 

$R_{net}=3R$

14. A battery of emf $E$ and internal resistance $r$ sends a current, $I_1,I_2$ when connected to an external resistance of $R_1,R_2$  respectively. Find the emf. and internal resistance of the battery.

Ans: It is given that,

$E$ is the emf of a battery

$r$ is the internal resistance which sends currents $I_1,I_2$

$R_1,R_2$ are external resistances

Current $I_1$ can be written as: $I_1=\frac{E}{R_1+r}$

$E=I_1(R_1+r)$  ……$(1)$

Similarly, $E=I_2(R_2+r)$ ……$(2)$

From equations $(1)$ and $(2)$

$\Rightarrow I_1(R_1+r)=I_2(R_2+r)$

$\Rightarrow I_{2}r-I_{1}r=I_1{R}_1-I_2{R}_2$

$\Rightarrow r(I_2-I_1)=I_1{R}_1-I_2{R}_2$

$\ast 35l\Rightarrow r=\frac{I_1{R}_1-I_2{R}_2}{I_2-I_1}$

Emf of the battery,$E$

$\Rightarrow E=I_1(R_1+r)$

$\Rightarrow E=I_1[R_1+\frac{{\text{I}}_1{\text{R}}_1-{\text{I}}_2{\text{R}}_2}{{\text{I}}_2-{\text{I}}_1}]$

$\Rightarrow E=I_1\left[\frac{{\text{I}}_2{\text{R}}_1-{\text{I}}_1{R}_1+I_1{\text{R}}_1-{\text{I}}_2{\text{R}}_2}{{\text{I}}_2-{\text{I}}_1}\right]$

$\Rightarrow E=\frac{I_1{I}_2({\text{R}}_1-{\text{R}}_2)}{{\text{I}}_2-{\text{I}}_1}$

Therefore, emf and internal resistance of the battery are $E=\frac{I_1{I}_2({\text{R}}_1-{\text{R}}_2)}{{\text{I}}_2-{\text{I}}_1}$ 

and 

$\ast 35lr=\frac{I_1{R}_1-I_2{R}_2}{I_2-I_1}$

15. Find the value of unknown resistance $X$ in the circuit shown in the figure if no current flows through the section AO. Also calculate the current drawn by the circuit from the battery of emf. $6V$ and negligible internal resistance. 

Ans: It is given that no current flows through AO then the circuit is said to be a balanced Wheatstone bridge. 

$\Rightarrow \frac{2}{4}=\frac{3}{X}$ 

$\Rightarrow X=\frac{12}{2}=6$

$\Rightarrow X=6\mathrm{\Omega }$

In branch AO, $$I=0$$

Resistance of $10\mathrm{\Omega }$ between A and O is ineffective and the circuit is reduced to:

$$2\mathrm{\Omega }$$ and $$4\mathrm{\Omega }$$ are in series; $$3\mathrm{\Omega }$$ and $$6\mathrm{\Omega }$$ are in series

6Ω and 9Ω are in parallel.

$\Rightarrow \frac{1}{R_P}=\frac{1}{6}+\frac{1}{9}=\frac{9+6}{54}=\frac{15}{54}$ 

$\Rightarrow R_P=\frac{54}{15}\mathrm{\Omega }$

$R_P$ and $$2.4\mathrm{\Omega }$$ are in series

$\Rightarrow R_{eff}=2.4+\frac{54}{15}$ 

$\Rightarrow R_{eff}=\frac{24}{10}+\frac{54}{15}=\frac{360+540}{150}=\frac{900}{150}$

$\Rightarrow R_{eff}=6\mathrm{\Omega }$

Current: $I=\frac{V}{R}=\frac{6}{6}=1$

$\Rightarrow I=1A$

Therefore, the unknown resistance $X=6\mathrm{\Omega }$ and current drawn by the circuit from the battery is $I=1A$.

16. 

(a) Obtain ohm’s law from the expression for electrical conductivity.

Ans: It is known that,

$I=neA{v}_d$

$J=\frac{I}{A}=ne{v}_d$

$V_d=\frac{eE\tau }{m}$

$\Rightarrow J=\frac{n{e}^{2}E\tau }{m}$ ; $J=\sigma E$

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

Let $l$ and $A$ be the length and area of the wire.

$I=JA$

$\Rightarrow I=\frac{n{e}^{2}E\tau }{m}\times A$ ; $(E=\frac{V}{l})$

$\Rightarrow I=\frac{n{e}^{2}V\tau }{ml}A$ 

$\Rightarrow V=\left(\frac{m}{n{e}^2\tau }\right)\left(\frac{l}{A}\right)I$ 

$\therefore V=RI$

$\Rightarrow R=\rho \frac{l}{A}$ where, $\rho =\frac{m}{n{e}^2\tau }$(Specific resistance of a wire)

(b) A cylindrical wire is stretched to increase its length by $10$ calculate the percentage increase in resistance?

Ans: It is given that; the length of cylindrical wire is stretched to increase its length by $10$.

Let, the original length be $l$ and the new length be $l^{\prime }$.

$\Rightarrow l^{\prime }=l+\frac{10}{100}l$ 

$\Rightarrow l^{\prime }=1.1l$

$\Rightarrow \frac{l^{\prime }}{l}=1.1$

As the volume of the wire remains the same

$Al=A^{\prime }{l}^{\prime }$

$\frac{A^{\prime }}{A}=\frac{l}{l^{\prime }}$

It is known that, $\text{R=}\rho \frac{l}{A}$

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

$\Rightarrow \frac{R^{\prime }}{R}=\frac{l^{\prime }}{A^{\prime }}\times \frac{A}{l}$

$$\Rightarrow \frac{R^{\prime }}{R}=\frac{l^{\prime }}{l}\times \frac{l^{\prime }}{l}={\left(\frac{l^{\prime }}{l}\right)}^2$$

$\Rightarrow \frac{R^{\prime }}{R}={(1.1)}^2=1.21$

Thus, the percentage increase in Resistance is $\frac{R^{\prime }-R}{R}\times 100=21$.

17. The current $I$ flows through a wire of radius $r$ and the free electron drift with a velocity ${\upsilon }_d$ what is the drift velocity of electrons through a wire of same material but having double the radius, when a current of $2I$ flows through it?

Ans: It is known that,

$I=neA{v}_d$

$\Rightarrow v_d=\frac{I}{neA}=\frac{I}{ne\pi r^2}$  ……$(1)$

If ${v_d}^{\prime }$ is the drift velocity of electrons in the second wire

$v{d}^{\prime }=\frac{I^{\prime }}{n{A}^{\prime }e}$ 

$\Rightarrow v{d}^{\prime }=\frac{2I}{n4\pi r^{2}e}$

$\Rightarrow v{d}^{\prime }=\frac{1}{2}(\frac{I}{n\pi r^{2}e})$   ……$(2)$

From equations $(1)$ and $(2)$

$\Rightarrow v{d}^{\prime }=\frac{vd}{2}$ 

Therefore, the drift velocity gets reduced to half.

18. Three identical cells, each of emf. $2V$ and unknown internal resistance are connected in parallel. This combination is connected to a $5$ ohm resistor. If the terminal voltage across the cell is $$\mathbf{1}.\mathbf{5}$$ volt. What is the internal resistance of each cell? Hence define the internal resistance of a cell? 

Ans: It is given that,

$$E=2V,V=1.5V,R=5\mathrm{\Omega }$$