by NCERT Solved Exercise Questions – Class 12 Physics Chapter 3 Current Electricity
3.1 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?
Ans – the emf of battery, E = 12 V
Internal resistance of battery, r = 0.4 Ohm
3.2 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?
Ans – The emf of the battery, E = 10 V
The internal resistance of the battery, r = 3 Ohm
Current in the circuit, I = 0.5 A
3.3 (a) Three resistors 1 Ω, 2 Ω, and 3 Ω 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.
Ans – (a) We are aware that when resistors are connected in series, their total resistance represents the effective resistance.
As a result, the three resistance combinations’ combined total resistance is 1 + 2 + 3 = 6 Ohm.
(b) Since the resistances are connected in series, the voltage/potential drop will vary, but the current through each of them will be equal to the current through the circuit.
according to Ohm’s law:
V = IR
12 = I x 6
I = 12/6 = 2 A
Now, using the same relation, voltage through resistors:
1 Ω : V(1) = 2 x 1 = 2V
2 Ω : V(2) = 2 x 2 = 4V
3 Ω : V(3) = 2 x 3 = 6V
(Note: V(1) + V(2) + V(3) = 2 + 4 + 6 = 12 V )
3.4 (a) Three resistors 2 Ω, 4 Ω and 5 Ω 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.
Ans –
2 Ω : I = 20/2 = 10 A
4 Ω : I = 20/4 = 5 A
5 Ω : I = 20/5 = 4 A
3.5 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10–4 °C–1 .
Ans – Given,
temperature coefficient of filament
3.6 A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10–7 m2 , and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?
Ans – length=15m
Resistance=5 ohm
3.7 A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
Ans – resistance r1 =2.1 ohm
resistance r2=2.7ohm
3.8 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 × 10–4 °C–1 .
Ans – The two quantities of current for the given voltage will equate to two distinct resistance values, which will lead to two different temperatures.
3.9 Determine the current in each branch of the network shown in Fig. 3.30:
Ans –
Let the circuit be like this
Where there are three current i1,i2,i3
3.10 (a) In a meter bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor S is of 12.5 Ω. Determine the resistance of R. 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 R and S 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?
Ans – l 1 = 39.5 cm at the balance point from the end A.
Y’s resistance is 12.5.
We are aware that the balance criteria for a metre bridge is:
3.11 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?
Ans – Emf E = 8 V
Internal resistance r = 0.5 Ω
The resistance of the resistor, R = 15.5 Ω
Let V’ be the effective voltage in the circuit.
Now, V’ = V – E
V’ = 120 – 8 = 112 V
Now, current flowing in the circuit is:
I = V’ / (R + r)
1215.5 + 0.5 = 7A
Now, using Ohm ’s Law:
Voltage across resistor R is v = IR
V = 7 x 15.5 = 108.5 V
Now, the voltage supplied, V = Terminal voltage of battery + V
Terminal voltage of battery = 120 -108.5 = 11.5 V
3.12 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?
Ans – In the second scenario, let E2 be the voltage.
The equilibrium condition is now provided by:
- 13 The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 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 × 10–6 m2 and it is carrying a current of 3.0 A.
Ans – We know,
vd :drift Velocity = length of wire(l) / time taken to cover
by substituting the given values
t = 2.7 x 104
electron to drift from one end of a wire to its other end is
3.14 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 × 106 m.)
Ans – Current over the entire globe = 1800 A
Radius of earth, r = 6.37 x 106 m
3.15 (a) Six lead-acid type of secondary cells each of emf 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?
Ans – Each cell’s emf, E = 2 V (In series)
Each cell’s intrinsic resistance, r = 0.015, (In series)
Additionally, the resistor’s resistance is R = 8.5.
Let I represent the circuit’s current draw.
3.16 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 × 10–8 Ω m, ρCu = 1.72 × 10–8 Ω m, Relative density of Al = 2.7, of Cu = 8.9.)
Ans –
Consequently, the aluminium wire is lighter for the same resistance and length.
Aluminum wire is utilised in power cables because it is lighter.
3.17 What conclusion can you draw from the following observations on a resistor made of alloy manganin?
| Current | Voltage | Current | Voltage |
| A | V | A | V |
| 0.2 | 3.94 | 3.0 | 59.2 |
| 0.4 | 78.7 | 4.0 | 78.8 |
| 0.6 | 11.8 | 5.0 | 98.6 |
| 0.8 | 15.7 | 6.0 | 118.5 |
| 1.0 | 19.7 | 7.0 | 138.2 |
| 2.0 | 39.4 | 8.0 | 158.0 |
Ans – The voltage to current ratio for the different numbers turns out to be fairly constant, or about 19.7.
As a result, the manganin alloy resistor adheres to Ohm’s law.
3.18 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?
Ans – (a) For a steady current flow, the conductor’s current remains constant.
Additionally, the area of the cross-section is inversely proportional to the current density, electric field, and drift speed. therefore not constant.
(b) Devices include semiconductor diodes, transistors, thermistors, thyristors, vacuum tubes, and more.
(c)According to Ohm’s law, V = I x R.
As a result, for a high value of current at a low voltage V, resistance R must be very low.
(d) If internal resistance is low, the supply could be harmed by the high current drawn during an unintentional short circuit.
3.19 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).
Ans – (a) Greater
(b) Lower
(c) Nearly independent
(d) 1022
3.20 (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) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?
(c) Determine the equivalent resistance of networks shown in Fig. 3.31.
Ans -(a)
(i) By connecting them in series, the greatest resistance can be achieved.
Rmax = R + R + R +……….n times = nR is the maximum resistance.
By mixing them in parallel with one another, it is possible to achieve the lowest effective resistance
(ii)To get minimum effective resistance, combine them in parallel. The effective resistance will be R/n.
(b)(ii)Connect 2 Ω and 3 Ω resistor in parallel and 1 Ω resistor in series to it
Equivalent Resistance R = {1/(1/2 + 1/3)} + 1 = 6/5 + 1
R = 11/5 Ω
(iii) 1 Ω+2 Ω+ 3 Ω= 6 Ω
, so we will combine the resistance in series.
(iv)Connect all three resistors in parallel.
Equivalent resistance is R = 1/(1/1 + 1/2 + 1/3) = (1x 2 x 3)/(6 + 3 + 2)
R = 6/11 Ω
3.21 Determine the current drawn from a 12V supply with internal resistance 0.5Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.
Ans – Now, we can write,
equivalent resistance = R’
‘ =[( R’)Parallel with (1)] + 1 + 1
3.22 Figure 3.33 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 kΩ 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 ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
(a) What is the value ε?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) 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?
(e) Would the circuit work well for determining 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?
Ans – (a) maintained constant emf of standard cell = 1.02V, balanced point of this cell = 67.3cm
Now when the standard cell is replaced by another cell with emf =
. balanced point for this cell = 82.3cm
Now as we know the relation
hence emf of another cell is 1.247V
(b) When the circuit is unbalanced, it only permits a modest current to pass through the galvanometer.
(c) No;
(d) No;
(e) No; and
(f) No. The circuit won’t function (E l).
By connecting a suitable resistor “R” in series with the wire AB, the circuit can be changed.
3.23 Figure 3.34 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.
Ans –
let the internal resistance of the cell be .
Now as we know, in a potentiometer
