Bihar Board Class 12 Physics Current Electricity Textbook Questions and Answers
Question 1.
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?
Answer:
Here, E = e.m.f of storage battery = 12 V
r = internal resistance = 0.4 Ω.
I = maximum current that can be drawn from the battery = ?
The maximum current is drawn from the battery when the external resistance in the circuit is zero i.e., R = 0.
Thus using the relation,
Question 2.
A battery of emf 10 V and internal resistance 3 Ω 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?
Answer:
Here, E = 10 V
r = 3 Ω
I = current in the circuit = 0.5 A.
R = resistance of the resistor = ?
V = terminal voltage of battery when circuit is closed = ?
Using the relestion I =
R =
or R = 17 Ω
Now using V = IR, we get
V = 0.5 x 17 = 8.5 A.
Question 3.
(a) Three resistors lfi, 1Ω,2Ω Rand 3Ω 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.
Answer:
Here,
R1 = 1Ω
R2 = 2Ω
R3 = 3Ω
Where R1, R2and R3 are the three registor connected in series.
(a) Let Rs = total resistance of the series combination = ?
In series, the total resistance is given by
Rs = R1 + R2 + R3 = 1 + 2 + 3 = 6 Ω.
.’. V1 = Potential drop across R1 = IR1 = 2 x 1 = 2V
V2 = Potential drop across R2 = IR2 = 2 x 2 = 4V
V3= Potential drop across R3 = IR3 = 2 x 3 = 6V
Question 4.
(a) (a) Three resistors 2Ω, 4Ω and 5Ω 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.
Answer:
Let R1, R2 and R3 be the three resistances connected in parallel.
Here,
R1 = 2 Ω
R2 = 4 Ω
R3 = 5 Ω
(b) Here, E = 20 V,r = 0
Let I = total current in the circuit = ?
and I1, I2,I1 be the current through R1R2 and R3, respectively?
∴ I1 = current through R1 =
I2 = current through R2 =
I3 = current through R3 =
Question 5.
At room temperature (27.0 °C) the resistance of a heating element is 100 12. What is the temperature of the element if the resistance is found to be 11712, given that the temperature coefficient of the material of the resistor is 1.70 x 10-4 °C-1.
Answer:
Let t°C = temperature of the element = ?
Here, Rt = Resistance at t°C = 117 Ω.
R27 = resistance at 27°C = 100 Ω.
α = temperature coefficient of resistance = 1.70 x 10-4 °C-1
Using the relation,
Question 6.
A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 x 10-7 m2, and its resistance is measured to be 5.01 Ω . What is the resistivity of the material at the temperature of the experiment?
Answer:
Here, l =. length of wire = 15 m.
A = area of cross-section of the wire = 6.0 x 10-7 m2.
R = resistance of wire = 5.012.
ρ = resistivity of the wire = ?
Using the relation, R = 
Question 7.
A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 oc. Determine the temperature coefficient of resistivity of silver.
Answer:
Here, R0 = R27.5 resistance of 27.5°C = 2.1 Ω
Rt = R100 = resistance at 100°C = 2.7 Ω
∝ = temperature coefficient of resistivity =?
∆t = 100 – 27.5 = 72.5°C.
Using the relation,
a = ,weget
Question 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 °C Temperature coefficient
of resistance of nichrome averaged over the temperature range involved is 1.70 x 10-4 °C-1
Answer:
Here, V = supply voltage = 230 V.
I1 = initial current 3.2 A
θ1 = initial temperature 27°C = room temperature
I2 = steady current = 2.8 A
θ2 = steady temperature =?
∝ = temperature coeïficient of resistance = 1.7
If R1 and R2, be the resistance of wire atθ1 and θ2 respectively, then
Answer:
Let I be the total current in the circuit.
I1 = Current flowing through AB.
.’. I – I1 = Current flowing through AD.
I2 – Current flowing throug BD.
.’. I1 – I2 = current flowing through BC. and I1 – I2 + I2 = Current flowing through DC.
Applying loop law to ABDA, we get 10 I1 + 5 I2 – 5 (I – I1) = 10
or 3I1 + I2 – I = 0 ……….(i)
Again applying loop law to BCDB, we get
5 (I1 -I2) -10 (I – I1 + I2) – 5I2 = 0
or 15I1 – 2I2 – 10I = 0
or 3I1 – 4I2 – 2I = 0 …….(ii)
Applying loop law to ABCEFA, we get
10I + 10I1 + 5 (I1 – I2) = 10
or 3F1 – I2 + 2I = 2 …………(iii)
(ii) + (iii) gives, 6I1 – 5I2 = 2 …….(iv)
Multiplying (i) by 2 and then adding to (iv) we get
9I1 + I2 = 2 ……..(v)
(vi) + 5 x (v) gives,
6I1 – 5I2 + 45I1 + 5I2 = 2 + 10
or 51 I1 = 12
or I1 =
∴ Current in branch AB, I1 =
∴ From (v) and (vi), we get
I2 = 2 – 9 x
– ve sign shows that I2 is actually from D to B. Now from (i), we get
Question 10.
(a) In a metre bridge [Fig.], the balance point is found to be at 39.5 cm from the end A, when the resistor Vis of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
(b) Determine the balance point of the bridge above if X and V are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?
Thick copper strips are used as they have negligible resistance, thus their use minimizes the resistance of connections so that the value of resistors of wheat stone/metre bridge are not changed.
(b) When X and Y are inter changed, then R = Y = 12.5 Ω, S = X = 8.16 0,l = ?
(c) Now A and C will be at the same potential so no current will flow through galvanometer. When at balance point galvanometer and cell are interchanged. No, it will not show any current.
Question 11.
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Answer:
Here, r = internal resistance of battery = 0.5 Ω
R = series register = 15.5 Ω
Ej = e.m.f. of storage battery = 8.0 V
E2 = Voltage of D.C. Supply= 120 V
As the storage battery of e.m.f. E1 is charged with a d.c. supply of e.m.f. E2, SO the net e.m.f. ‘E’ in the circuit is given by
E = E2 – E1 = 120 – 8 = 112 V
Total resistance in the circuit, R’ = R + r = 15.5 + 0.5 = 16.0 Ω.
Let V = terminal voltage of the battery during charging = ?
If I = current in the circuit during charging, then
I =
During charging, the voltage of the D.C. supply in a circuit must be equal. to the sum of the voltage drop across R and terminal voltage of the battery.
i.e., E2 = V + IR
or 120 = V + 7 x 15.5
or V = 120 – 108.5 = 11.5 V
V can also be calculated as :
V = sum of e.m.f. of battery + P.D. across r = E + Ir = 8 + 7 x 0.5 = 11.5 V.
The purpose of having a series register in the charging circuit is to limit the current drawn from the external source of d.c. supply which is dangerously high in its absence.
Question 12.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Answer:
Here,
E1 = e.m.f of first cell = 125 V
l1 = balancing length for first cell 35 cm
E2 = e.m.f of 2nd cell = ?
l2 = balancing length for second cell = 63 cm
Using the relation,
E2 =
= 9 x 0.25 = 2.25 V.
Question 13.
The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 x 1028 m– 3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 x 10-6 m2 and it is carrying a current of 3.0 A.
n = 8.5 x 1028 m-3
Answer:
Here, n = number density of free electrons = 8.5 x 1028 m-3
l = length of wire = 3 m
A = Area of cross-section of wire = 2.0 x 10-6 m2
I = current in the wire = 3.0 A
e = 1.6 x 10-19 C
Let t = time taken by electron to drift from one end to another of the wire = ?
Using the relation, I = neA υd , we get
vd = I/neA
Question 14.
The earth’s surface has a negative surface charge density of 10-9 C m-2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 x 106 m).
Answer:
Here, a = surface charge density of earth = 10-9 cm2
R = radius of earth = 6.37 x 106 m
I = Current over entire globe = 1800 A
V = P.D. between the top of atmosphere and surface of earth = 4000 KV.
Let t = time required to neutralise earth’s surface = ?
∴ Area of whole globe, A = 4πR2
= 4π x (6.37 x 106)2 = 509.64 x 1012 m2.
Also we know that
σ =
t =
= 0.283 x 103 S
= 283 s = 4 min 43 s.
Question 15.
(a) Six lead-acid type of secondary cells each of tmf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?
(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?
Answer:
Here, e.m.f. of each secondary cell, E = 2.0 V
n = no. of cells = 6
r = internal resistance of each cell = 0.015 Ω
R = external resistance = 8.5 W
Total e.m.f = nE = 6 x 2 = 12 V R + nr = 8.5 + 6 x 0.015 = 8.59 Ω.
(i) If I be the total current, then
(ii) Let V = terminal voltage = IR = 1.4 x 8.50 = 11.9 V
V can also be calculated as :
V = nE – Im = 12-1.4 x 0.015 x 6 = 11.876 = 11.9 V
(b) Here, E = e.m.f. of cell = 1.9 V.
r = internal resistance of cell = 380 Ω.
I max= maximum current that can be drawn from the cell = ?
We know that I =
For I = Imax , R must be zero.
Imax =
No, the cell cannot be used to drive the starting motor of a car as initially it needs a current of nearly 100 A.
Question 16.
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (ρal = 2.63 x 10-8 m, ρCu = 1.72 x 10-8 Ω m, Relative density of Al = 2.7, of Cu = 8.9).
Answer:
Let A1 A2 be the area of cross-section of aluminium and copper wires respectively.
l = lenth of each wire
Also let R1, R2 be the resistance of aluminium and copper wires respectively. .
R1 = R2 (given) ……(i)
For Aluminium Wire ρ2 = resistivity of Al = 2.63 x 108 Ωm.
relative density of Al = 2.7
∴ Its density, d1 = 2.7 x 103 kg m-3.
Thus using the relation,
R = ρ
or R = 2.63 x 10-8 x
if m1 = mass of al wire then
m1 = density x volume
= d1 x A1l1 = 2.7 x 103 x A1 l ………(iii)
p =resistivity of Cu = 1.72 x 10-8Ωm.
relative densitv of Cu = 8.9
∴ Its density, d2 = 8.9 x 103 kg m-3.
∴ R2 =
m2 = mass of Cu wire = A2d2,P2
= A2 x 8.9 x 103 x I ………..(v)
From (i), (ii) and (iv), we get
It follows from (vii) that the Aluminium wire of given length is lighter than copper wire of same length since for the same values of resistance and length, the aluminium wire has lesser mass than copper wire, so aluminium wire is preferred for overhead power cables. A heavy cable may sag down owing to its own weight.
Question 17.
What conclusion can you draw from following observations on a resistor made of alloy manganin:
Answer:
Ohm’s law is valid to a high accuracy.
We know that R =
From the observations given here, for all currents from 0.2 A to 8.0 A, the resistance of the registor is almost same and is equal to 19.7 Q. As the current increases, temperature also increases but resistance has no effect. Also we know that the resistance of alloys i.e., manganin here does not change with temperature and their temperature coefficient of resistance is negligibly small. Thus the resistance as well as resistivity of the alloy manganin is nearly independent of temperature.
Question 18.
Answer the following questions:
(a) A steady current flows in a metallic conductor of non-uniform cross- section. Which of these quantities is constant along the conductor current, current density, electric field, drift speed?
(b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.
(c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?
(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why? ‘
Answer:
(a) We know that current density, electric field and drift speed all are inversely proportional to the area of cross-section of the conductor. As the steady current flows in the metallic conductor of non-uniform area of cross-section, so only current remains constant, when it flows through a conductor of non-uniform area of cross-section.
(b) No, Ohm’s law is not universally applicable for all conducting elements. It is valid only for metallic conductors and that also when the physical conditions like temperature, stress etc. of the conductor remains the same.
The examples of elements which don’t obey Ohm’s law are vacuum diode semi conductor, diodes, thermistors, thyristor SCR etc.
(c) The maximum current that can be drawn from a voltage supply is given by
Imax =
Where E is the e.m.f. of the source and r is the internal resistance of the source. Clearly for Imax to be large, r must be small.
(d) If tire internal resistance (r) of the HT supply is low, large amount of current will be drawn from the supply during the short circuiting which will cross the safety limits and damage the HT source. But if r is large, then current in the circuit will not exceed the safe limit. Hence a high voltage supply must have a very large internal resistance.
Question 19.
Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature^
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (1022/1023).
Answer:
(a) greater
(b) lower
(c) nearly independent of
(d) 1022
Question 20.
(a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?
(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i)
(c) Determine the equivalent resistance of networks shown in Figure.
Answer:
(a) (i) The resistance will be maximum when the resistor are connected in series. Thus if Rmax be the maximum resistance, then
Rmax = R + R + R + …… + n times = nR.
(ii) The resistance will be minimum when the resistors are connected in parallel thus if Rmax be the minimum resistance, then
(b) (i) When 3 Ω resistance is connected in series with the parallel combination of 1 and 2 Ω, we will get the required resistance.
(ii) When 1 Ω resistance is connected in series with the parallel combination of 2 Ω and 3 Ω, we will get the required resistance.
(iv) When all the resistors are connected in parallel, then
Answer:
(c) (a) The given network is a series combination of four equal units. Each unit has 4 resistances in which two resistances (1Ω. each being in series) are in parallel with 2 resistances (2 Ω each in series) If R be the net resistance of one unit, then
If R be the total resistance of the network, then
R = 4 x Rp = 4 x
(b) Let us connect a battery across a battery across the points A and B. We see that the same current flows through all the five resistances as all the 5 resistances are connected in series.
If R’ be the total resistance of the network between the point A and B, then
R’ = 5R.
Question 21.
Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by infinite network shown in figure. Each resistor has resistance of 1Ω.
Answer:
Let X be the equivalent resistance of the network. Since network is infinite adding one more set of three resistances each of value R = 1 Ω across the terminals will not affect the total resistance i.e., it should still remain equal to X. Thus this network can be represented as
Let Req be the equivalent resistance of this network, then
Req = R + equivalent resistance of parallel combination of X and R) + R
Addition of 3 resistances to resistance X of infinite network should not alter the total resistance of the infinite network. Thus
= R +
= 2R +
Addition of 3 resistances to resistance X of infinite network should not alter the total resistance of the infinite network. Thus
Req = X
Question 22.
Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 k Ω is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf e and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
(a) What is the value el
(b) What purpose does the high resistance of 600 k have?
(c) Is the balance point affects by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?
(f) Would the circuit work well for determing an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?
Answer:
Here, E1 = e.m.f. of standard cell = 1.02 V. ”
E2 = e.m.f. cell connected = e = ?
l1 = balancing length for E1 = 67.3 cm
l2 = balancing length for E2 = 82.3 cm.
E = e.m.f. of driver cell = 2.0 V, r = its internal resistance = 0.40 Ω.
(a) As =
∴ ε = E1 x
(b) Here R = 600 K Ω connected in series with cell. The purpose of using this high resistance is to allow very small current through the galvanometer when the movable contact is far from the balance point.
(c) No, the balance point is not affected by the presence of this high resistance.
(d) No, the balance point is not affected by the internal resistance of the driver cell.
(e) No, the method will not work as the balance point will not be obtained on the potentiometer wire if the e.m.f. of the driver cell is less than the e.m.f. of the standard cell.
(f) No, the circuit will not work for measuring extremely small e.m.f. of the order of milli vilt because the balance point will be just close to the end A and the percentage error in measurement will be very large. The circuit is modified by putting a suitable resistance R in series with the wire AB so that potential drop across AB is only slightly greater than the emf. to be measured. Then, the balance point will be at larger length of the wire and the percentage error will be much smaller.
Question 23.
Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε?
Answer:
Here, R = 10 Ω, X = ?
l1 = balancing length with standard resistance R = 58.3 cm
l2 = balancing length with standard resistance X = 68.5 cm
For a potential we know that
Also we know that
In case we don’t get balance point on potentiometer wire, e.m.f. of driver cell is less than the e.m.f to be measured. Then we need driver cell of higher e.m.f. When we don’t get balance point, it also means that the potential drop across R or X is greater than the potential drop across wire AB. A suitable\series resistor can be put in the external circuit to reduce the current in the outside circuit and hence potential drop across Ror X.
Question 24.
Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
Answer:
Here, e.m.f. of driver cell, E = 2.0 V
e.m.f. standard cell, E1 = 1.5 V
r = internal resistance of cell = ?
l1= balancing length when cell is in open circuit = 76.3 cm
l2 = balancing length when cell in closed circuit = 64.8 cm.
R = Resistance used = 9.5 Ω
We know that