NCERT Exemplar Class 12 Physics Solutions Chapter 3 Current Electricity

Question:2

Two batteries of emf $\varepsilon _{1}$ and $\varepsilon _{2}\left ( \varepsilon_ {2}>\varepsilon _{1} \right )$and internal resistances r₁ and r₂ respectively are connected in parallel as shown in Figure.


A. The equivalent emf $\varepsilon _{eq}$ of the two cells is between $\varepsilon _{1}$ and $\varepsilon _{2}$, i.e. $\varepsilon _{1}<\varepsilon _{eq}<\varepsilon _{2}$
B. The equivalent emf $\varepsilon _{eq}$ is smaller than 1.
C. The $\varepsilon _{eq}$ is given by $\varepsilon _{eq}=\varepsilon _{1}+ \varepsilon _{2}$ always.
D. $\varepsilon _{eq}$ is independent of internal resistances r₁ and r₂.

Answer:

The correct answer is the option (Q)
Explanation: Equivalent EMF in parallel is given as:
$\varepsilon _{eq}=\frac{\varepsilon _{1}r_{2}+\varepsilon _{2}r_{1}}{r_{1}+r_{2}}$
This value will always lie between $\varepsilon _{1}$ and $\varepsilon _{2}$

Question:3

A resistance R is to be measured using a meter bridge. Student chooses the standard resistance S to be 100Ω. He finds the null point at $l_1 = 2.9 cm$. He is told to attempt to improve the accuracy. Which of the following is a useful way?

A. He should measure $l_1$ more accurately.
B. He should change S to 1000Ω and repeat the experiment.
C. He should change S to 3Ω and repeat the experiment.
D. He should give up hope of a more accurate measurement with a meter bridge.

Answer:

The correct answer is the option (C)
Explanation:
$R=S\left (\frac{l_1}{100-l_1} \right )=100\left (\frac{2.9}{97.1} \right )=2.98\Omega$
He should aim to get the balance point near the centre (l1=50 cm)
$2.98\Omega =S\left (\frac{50}{50} \right )$
$S\cong 3\Omega$

Question:4

Two cells of emf’s approximately 5V and 10V are to be accurately compared using a potentiometer of length 400cm.
A. The battery that runs the potentiometer should have a voltage of 8V.
B. The battery of potentiometer can have a voltage of 15V and R adjusted so that the potential drop across the wire slightly exceeds 10V.
C. The first portion of 50 cm of the wire itself should have a potential drop of 10V.
D. Potentiometer is usually used for comparing resistances and not voltages.

Answer:

The correct answer is the option (B)
Explanation: The potential drop along the wires of the potentiometer should be greater than the emfs of cells. The battery of potentiometer must be in excess of 10V.

Question:5

A metal rod of length 10 cm and a rectangular cross-section of $1cm \times \frac{1}{2} cm$ is connected to a battery across opposite faces. The resistance will be
A. maximum when the battery is connected across $1cm \times \frac{1}{2} cm$ faces.
B. maximum when the battery is connected across $10cm \times 1 cm$ faces.
C. maximum when the battery is connected across $10cm \times \frac{1}{2} cm$faces.
D. same irrespective of the three faces.

Answer:

The correct answer is the option (A)
Explanation:$R=\rho\frac{l}{A}$
Resistance is directly proportional to length and inversely proportional to the cross-sectional area. Resistance will be maximum when l is maximum and A minimum. Both these conditions are met simultaneously when the battery is connected across the $1cm \times \frac{1}{2} cm$ faces.

Question:6

Which of the following characteristics of electrons determines the current in a conductor?
A. Drift velocity alone.
B. Thermal velocity alone.
C. Both drift velocity and thermal velocity.
D. Neither drift nor thermal velocity

Answer:

The correct answer is the option (a)
$I=Anev_d$

Question:8

Consider a simple circuit shown in Figure variable resistance R’. R’ can vary from $R_0$ to infinity. r is internal resistance of the battery $(r<<R<<R_0).$

A. Potential drop across AB is nearly constant as R’ is varied.
B. Current through R is nearly a constant as R’ is varied.
C. Current I depends sensitively on R’.
D.$I\geqslant \frac{V}{r+R}$ always.

Answer:

The correct answers are the options (A,D)
Explanation:
$I=\frac{V}{r+R_{eq}}$
As $r+R<r+R_{eq}$
$I=\frac{V}{r+R_{eq}}\leq \frac{V}{r+R}$

Question:9

Temperature dependence of resistivity $\rho (T)$ of semiconductors, insulators and metals is significantly based on the following factors:
A. number of charge carriers can change with temperature T.
B. time interval between two successive collisions can depend on T.
C. length of material can be a function of T.
D. mass of carriers is a function of T.

Answer:

The correct answers are the options (A,B)
Explanation: Resistivity is a function of relaxation time$(\tau )$ and mass of charge carrier (m). Mass of charge carrier is independent of temperature, whereas length also does not fluctuate significantly over a temperature range.

Question:10

The measurement of an unknown resistance R is to be carried out using Wheat stones bridge (see Figure of NCERT Book). Two students perform an experiment in two ways. The first students take $R_2 = 10\Omega$ and $R_1 = 5\Omega$. The other student takes $R_2 = 1000\Omega$and $R_1 = 500\Omega$. In the standard arm, both take $R_3= 5\Omega$. Both find $R=\frac{R_{2}}{R_{1}}R_{3}=10\Omega$ within errors.

A. The errors of measurement of the two students are the same.
B. Errors of measurement do depend on the accuracy with which $R_{2}$ and $R_{1}$ can be measured.
C. If the student uses large values of $R_{2}$ and $R_{1}$, the currents through the arms will be feeble. This will make determination of null point accurately more difficult.
D. Wheatstone bridge is a very accurate instrument and has no errors of measurement.

Answer:

The correct answers are the options (b,c)
Explanation: Using large values of $R_{1}$ and $R_{2}$ , will reduce the current flowing through the system, making it difficult to get accurate readings from the galvanometer.

Question:11

In a meter bridge the point D is a neutral point (Figure).


A. The meter bridge can have no other neutral point for this set of resistances.
B. When the jockey contacts a point on meter wire left of D, current flows to B from the wire.
C. When the jockey contacts a point on the meter wire to the right of D, current flows from B to the wire through galvanometer.
D. When R is increased, the neutral point shifts to left.

Answer:

The correct answers are the options (a,c)
Explanation: No current will flow through the Galvanometer when the jockey is placed at D. Potential Difference between B and D is zero. If the jockey is placed to the right of D, Potential at B becomes greater than the potential at D, the current flows from B to D.

Question:13

The relaxation time $\tau$ is nearly independent of applied E field whereas it changes significantly with temperature T. First fact is (in part) responsible for Ohm’s law whereas the second fact leads to variation of ρ with temperature. Elaborate why?

Answer:

Relaxation time decreases with increase in drift velocity. This increases the resistivity
$\rho=\frac{m}{ne^2\tau }$

Question:14

What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate $R_{unknown}$ by any other method?

Answer:

As no electricity will pass through the galvanometer of the Wheatstone bridge at the null point. It can be used to calculate the resistance of an unknown resistor.

Question:15

What is the advantage of using thick metallic strips to join wires in a potentiometer?

Answer:

Thick metallic strips have large cross section and thus offer less resistance

Question:16

For wiring in the home, one uses Cu wires or Al wires. What considerations are involved in this?

Answer:

The following considerations are made to decide on the wiring:

• Conductivity of metal
• Cost
• Availability

Question:17

Why are alloys used for making standard resistance coils?

Answer:

Because of their low temperature sensitivity and low temperature coefficient of resistance, alloys are used to make standard resistance coils. Also, they have a lower conductivity or higher resistivity, this reduces the length required to make coils.

Question:18

Power P is to be delivered to a device via transmission cables having resistance $R_C$. If V is the voltage across R and I the current through it, find the power wasted and how can it be reduced.

Answer:

P (Power transmitted)=VI
H (Heat Loss)=$I^{2}R$
High current flowing through the devices will lead to higher Energy Loss (as heat), so power should be transmitted at a high voltage and low current.

Question:19

AB is a potentiometer wire (Figure). If the value of R is increased, in which direction will the balance point J shift?

Answer:

On increasing the applied resistance, the potential difference across AB will decrease. To balance the potential across the other battery, the length AJ' should increase. Balance point shift towards B.

Question:20

While doing an experiment with potentiometer (Figure ) it was found that the deflection is one sided and (i) the deflection decreased while moving from one end A of the wire to the end B; (ii) the deflection increased. While the jockey was moved towards the end B.


(i) Which terminal +or –ve of the cell$E_{1}$, is connected at X in case (i) and how is $E_{1}$ related to $E$?
(ii) Which terminal of the cell $E_{1}$ is connected at X in case (ii)?

Answer:

i) As jockey moves closer to B, the galvanometer potential difference decreases. This requires the positive terminal of $E_{1}$ to be at X.
ii) Exactly opposite to (i), at X $E_{1}^{'}$s negative terminal is connected.

Question:21

A cell of emf E and internal resistance r is connected across an external resistance R. Plot a graph showing the variation of P.D. across R, verses R.

Answer:

$I=\frac{E}{R+r} and V=IR$
$V=\frac{ER}{R+r}=\frac{E}{1+\frac{r}{}R}$
An increase in R leads to an increase in V, till the point it approaches E
Following is the graphical representation:

Question:23

Let there be n resistors $R_1 ............R_n$ with $R_{max} = max (R_1......... R_n)$ and $R_{min} = min (R_1 ..... R_n)$. Show that when they are connected in parallel, the resultant resistance R_P< Rmin and when they are connected in series, the resultant resistance $R_S > R_{max}$. Interpret the result physically.

Answer:

$\frac{1}{R_{parallel}}=\frac{1}{R_1}+\frac{1}{R_{2}}+.....+\frac{1}{R_n}$
$\frac{R_{min}}{R_{parallel}}=\frac{R_{min}}{R_1}+\frac{R_{min}}{R_2}....+\frac{R_{min}}{R_n}$
$\frac{R_{min}}{R_{parallel}}> 1$
$R_{series}= R_1+R_2...+R_n$
$R_{series}>R_{max}$
This can be represented physically as shown below:

Question:24

The circuit in Figure shows two cells connected in opposition to each other. Cell $E_{1}$ is of emf 6V and internal resistance 2$\Omega$; the cell $E_{2}$ is of emf 4V and internal resistance 8$\Omega$. Find the potential difference between the points A and B.

Answer:

Equivalent EMF of the Cells $=6V-4V=2V$
$R_{eq}=2\Omega +8\Omega=10\Omega$
$I=\frac{2V}{10 \Omega}=0.2A$
$V_B-V_A=4+8\times 0.2=5.6V$

Question:25

Two cells of same emf E but internal resistance $r_{1}$ and $r_{2}$ are connected in series to an external resistor R (Figure). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero.

Answer:

$EMF_{ef}=E+E=2E$
$R_{ef}=r_1+r_2$
$I=\frac{2E}{R+r_1+r_2}$
As net potential across the cell is 0,
$E-Ir_1=0$
$E-\frac{2Er_1}{R+r_1+r_2}=0$
$R=r_1-r_2$

Question:26

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1mm. Conductor B is a hollow tube of outer diameter 2mm and inner diameter 1mm. Find the ratio of resistance to

Answer:

$Resistance(R)=\rho\frac{L}{A}$
$R_A=\rho _A\left (\frac{L_A}{A_A} \right ) and R_B=\rho _B\left (\frac{L_B}{A_B} \right )$
$\rho_A=\rho_B$ and $L_A=L_B$
$\frac{R_A}{R_B}=\frac{A_B}{A_A}=\frac{\pi r_A^2}{\pi (r_{out}^2-r_{in}^2)}=\frac{1^2-\left (\frac{1}{2} \right )^{2}}{\left ( \frac{1}{2} \right )^{2}}=\frac{\frac{3}{4}}{\frac{1}{4}}=3$

Question:27

Suppose there is a circuit consisting of only resistances and batteries and we have to double (or increase it to n-times) all voltages and all resistances. Show that currents are unaltered. Do this for circuit of Example 3.7 in the NCERT Text Book for Class XII.

Answer:

Original:
$EMF = V_1$
$Resistance = R + R_{eff}$
$Current = \frac{V_1}{R+R_{eff}}$
New:
$EMF = nV_1$
$Resistance =n( R + R_{eff})$
$Current =\frac{n(V_1)}{n(R+R_{eff})}= \frac{V_1}{R+R_{eff}}$
Increasing the EMF and Resistance in the same ratio (say 2), will not change the current flowing through the system.

Question:29

A room has AC run for 5 hours a day at a voltage of 220V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions?

$[\rho_ {Cu} = 1.7 \times -810\Omega m, \rho _{Al} = 2.7 \times 10^{-8} \Omega m]$

Answer:

Energy consumed day=10 kWh
Energy consumed hour=2 kWh
Power=2kW
P=VI
V=220V
$I=\frac{P}{}V=\frac{2000}{220}=\frac{100}{11}$
Heat $=I^2\times R_{Cu}$
$R_{Cu}=\rho\left ( \frac{L}{}A \right )=1.7\times 10^{-8}\times \left (\frac{10}{\pi \times \left (5\times 10^{-4} \right )^{2}} \right )$
$Heat=\left (\frac{100}{110} \right )^{2}\times 1.7\times 10^{-8}\times \left (\frac{10}{\pi \times \left (5\times 10^{-4} \right )^{2}} \right )=4.4 W$
fractional loss in heating
$=\frac{4.4}{2000}\times 100 =0.22\%$
$\frac{Heat_{Al}}{Heat_{Cu}}=\frac{I^2RAl}{I^2R_{Cu}}=\frac{\rho _{Al}}{\rho _{Cu }}$
Heat_Al=7W
fractional loss in heating
$=\frac{7}{2000}\times1000=0.35\%$

Question:30

In an experiment with a potentiometer, $V_{B}=10V$ R is adjusted to be $50 \Omega$ (Figure). A student wanting to measure voltage E₁ of a battery (approx. 8V) finds no null point possible. He then diminishes R to $10 \Omega$ and is able to locate the null point on the last (4^th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

Answer:

R_eq=50+R
V_eq=10V
$I=\frac{10}{50+R'}$
The potential difference across the potentiometer
$V'=IR'=\frac{10R'}{50+R'}$
V'<8
$10R'<8(50+R' )$
R'<200
$\frac{10R'}{10+R'}>8$
R'>40
At balance point:
$\frac{10\times \frac{3}{}4R'}{10+R'}<8$
R'>-160
As R’ can only be positive, R'>160
$k\times 400cm>8V$
k>2 V/m
As balance, point isn’t at 3m
$K\times 3<8$
$k<2\frac{2}{}3\frac{V}{}m$

Question:31

(a) Consider the circuit in Figure. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity?

(b) Electrons give up energy at the rate of $RI_2$ per second to the thermal energy. What time scale would one associate with energy in problem (a)? n = no of electron/volume = $10^{29}/m^3$, length of circuit = 10 cm, cross-section = A $= (1mm)^2$

Answer:

a)
$A=10^{-6}m^2$
$R=6\Omega$
$n=10^{29} m^{-3}$
$e=1.6\times 10^{-19}C$
$I=\frac{V}{}R=1A$
$m_e=9.1\times10^{-31}Kg$
$L=10^{-1}m$
a) $I=Anev_d$
$v_d=\frac{I}{Ane}=6.25\times 10^{-5}m/s$
Kinetic energy=
$\frac{1}{}2\times m_e\times v_d^2=2\times 10^{-17}J$
b) Power loss=
$I^2R=6W$
$P=\frac{E}{}t$
$t=\frac{E}{}P=10^{-17}s$