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### Chapter 3 Current Electricity

## Table of Contents

### Answer The Following Question Answer

**Question****1.The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery?**

**Solution :**

Given that Emf of the battery, E = 12 V

Battery has an Internal resistance of R = 0.4 Ω

The Maximum current drawn from the battery is given by = I

According to Ohm’s law,

E = IR

I=E / R I=12/0.4 = 30 A

The maximum current drawn from the given battery is 30 A.

**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?**

**Solution :**Electromotive force (E) of battery = 10 volt.

Internal Resistance r of the battery = 3 Ω

Current in the Circuit (I) = 0.5 A

Let the resistance of the required resistor is R Ω

Let the Terminal voltage of the register be V.

As per Ω ’s law,

.

**Question****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.**

**Solution :****The diagram is given as:**

A) When three resistors are connected in series then the total resistance is equal to sum of individual resistance.

Total resistance is equal to = 1 + 2 + 3 = 6 Ω

(b) Let the current flowing in the circuit be I.

Electromotive force EMF, (E) = 12V

Total resistance in the circuit R = 6 Ω

According to Ohm ’s law, V = IR

**Question****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.**

**Solution :**

.

(b)

**Question****5.****At room temperature (27.0 °C) the resistance of a heating element is 100Ω. What is the temperature of the element if the resistance is found to be 117Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10 ^{–4} °C^{–1}.**

**Solution :**Given: Temperature coefficient of filament,

α = 1.70 × 10^{-4} per °C

Let T_{1} be the temperature of element, R_{1 }= 100Ω (Given: T_{1 }= 27°C)

Let T_{2} be the temperature of element, R_{2 }= 117Ω

To find T_{2 }= ?

The formula is: R_{2} = R_{1}[1 + α(T_{2}-T_{1})]

⇒ R_{2 }– R_{1 }= R_{1}α(T_{2 }– T_{1})

⇒ R_{2 }– R_{1 }= R_{1}α (T_{2 }– T_{1})

**Question****6. A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10 ^{−7 }m^{2}, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?**

**Solution :**Given, the length of wire l = 15 metre

Area of cross section of wire a = 6.0 × 10^{-7} metre square

Resistance of material of wire,R = 5 Ω

Resistivity of material of wire = ρ

.

**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 °C. Determine the temperature coefficient of resistivity of silver.**

**Solution :**Resistance of silver wire at 27.5 °C R_{1} = 2.1 Ω

Resistance of silver wire at 100 ° C R_{2} = 2.7 Ω

Let temperature coefficient of silver wire is given by α.

The formula for α is: R_{2} = R_{1}[1 + α(T_{2}-T_{1})]

⇒ R_{2 }– R_{1 }= R_{1}α(T_{2 }– T_{1})

⇒ R_{2 }– R_{1 }= R_{1}α (T_{2 }– T_{1})

**Question****8. Aheating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds toa 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}.**

**Solution :**Given Supply Voltage = 230 Volt

And it draws current I_{1} = 3.2 Ampere

Resistance is given by R_{1} = V/I

R_{1} = 230V/3.2A

R_{1} = 71.87 Ω

Steady value of current I_{2} = 2.8Ampere

Resistance R α is given by R_{2} = 230V/2.8A

R_{2} = 82.14 Ω

Temperature coefficient of nichrome wire α = 1.70 × 10^{-4} per °C

Initial temperature of nichrome T_{1} = 27 ° C

Steady Temperature attained by nichrome = T_{2}

The formula is: R_{2} = R_{1}[1 + α(T_{2}-T_{1})]

⇒ R_{2 }– R_{1 }= R_{1}α(T_{2 }– T_{1})

⇒ R_{2 }– R_{1 }= R_{1}α (T_{2 }– T_{1})

**Question****9. Determine the current in each branch of the network shown in fig 3.30:**

**Solution :The current through the labelled diagram shows the current flowing the respective branches:**

For the closed circuit ABDA, potential is zero i.e.,

10I_{2} + 5I_{4} − 5I_{3} = 0

2I_{2} + I_{4} −I_{3} = 0

I_{3} = 2I_{2} + I_{4} … (1)

For the closed circuit BCDB, potential is zero i.e.,

5(I_{2} − I_{4}) − 10(I_{3} + I_{4}) − 5I4 = 0

5I_{2} + 5I_{4} − 10I_{3} − 10I_{4} − 5I_{4} = 0

5I_{2} − 10I_{3} − 20I_{4} = 0

I_{2} = 2I_{3} + 4I_{4} … (2)

For the closed circuit ABCFEA, potential is zero i.e.,

−10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0

10 = 15I_{2} + 10I_{1} − 5I_{4}

3I_{2} + 2I_{1} − I_{4} = 2 … (3)

From equations (1) and (2), we obtain

I_{3} = 2(2I_{3} + 4I_{4}) + I_{4}

I_{3} = 4I_{3} + 8I_{4} + I_{4}

− 3I_{3} = 9I_{4}

− 3I_{4} = + I_{3} … (4)

Putting equation (4) in equation (1), we obtain

I_{3} = 2I_{2} + I_{4}

− 4I_{4} = 2I_{2}

I_{2} = − 2I_{4} … (5)

It is evident from the given figure that,

I_{1} = I_{3} + I_{2} … (6)

Putting equation (6) in equation (1), we obtain

3I_{2} + 2(I_{3} + I_{2}) − I_{4} = 2

5I_{2} + 2I_{3} − I_{4} = 2 … (7)

Putting equations (4) and (5) in equation (7), we obtain

5(−2 I_{4}) + 2(− 3 I_{4}) − I_{4} = 2

− 10I_{4} − 6I_{4} − I_{4} = 2

17I_{4} = − 2

I_{4} = -2/17Ampere

Equation (4) reduces to

I_{3} = − 3(I_{4})

I_{3} = -3 × -2/17

I_{3} = 6/17Ampere

I_{2} = -2(I4)

I_{2} = -2 × -2/17

I_{2} = 4/17Ampere

I_{2 }– I_{4} = 6/17Ampere

I_{3} + I_{4} = 6/17Ampere

I_{1} = I_{2} + I_{3} = 4/17 + 6/17 = 10/17 Ampere

Therefore, current in branch AB = 4/17Ampere

In branch BC = 6/17Ampere

In branch CD = -4/17Ampere

In branch AD = 6/17Ampere

In branch BD = -2/17Ampere

Total current = 4/17 + 6/17 + -4/17 + 6/17 + -2/17 = 10/17Ampere.

**Question****10. (a) In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor Y is 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 Y 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?**

**Solution :The diagram is shown below:**

(b) If X and Y are interchanged, then l_{1} and 100−l_{1} get interchanged.

The balance point of the bridge will be 100−l_{1} from A.

100−l_{1 }= 100 − 39.5 = 60.5 cm

Therefore, the balance point is 60.5 cm from A.

(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection. Hence, no current would flow through the galvanometer.

**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?**

**Solution :**

Electromotive force EMF of battery = 8Volt

Internal resistance of battery r = 0.5 Ω

Supply Voltage V = 120Volt

Resistance of resistor R = 15.5 Ω

Effective voltage in circuit = V1

Since Resistance R is connected in series Hence We can write V1 = V-E

V_{1} = 120V-8V = 112 V

Current flowing in the circuit:

.

**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?**

**Solution :**Given EMF of the cell E1 = 1.25Volt

Balance point of the potentiometer l1 is given 35cm

Since the cell is replaced by another cell of EMF E2

New balance point of potentiometer l2 is 63cm

.

**Question****13. The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10 ^{28} 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 }m^{2} and it is carrying a current of 3.0 A.**

**Solution :**Number density of free electrons in a copper conductor, n = 8.5 × 10^28 per meter cube

Length of Copper Wire l = 3 meter

Area of cross section of wire A = 2 × 10^{-6} meter square

Current drawn by wire is given 3 Ampere

Since I = nAeV_{d}

Here V_{d} is the drift velocity and “e” is the electron charge 1.6 × 10^-19Coulomb.

**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 × 10^{6} m.)**

**Solution :**Surface Charge density of earth d = 10^{-9} Coulomb per meter square

Current over the entire globe = 1800 Ampere

Radius of earth r = 6.37 × 10^{6} meter

Surface area of earth A = 4×π×radius×radius

A = 4×π×6.37×10^{6}×6.37×10^{6}

A = 5.09 × 10^{14} m^{2}

Charge on the earth surface q = d × A

q = 10^{-9} × 5.09 ×10^{14} m^{2}

q = 5.09 × 10^{5} Coulomb

Let the time taken to neutralize earth surface = t

**Question****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?**

**Solution :**A) Number of secondary cells n = 6

Emf of each secondary cell E = 2Volt

Internal resistance of each cell r = 0.015 Ω

Resistance of resistor R = 8.5 Ω

Let current drawn from supply = I

Hence maximum current drawn from cell is 0.005Ampere and since a large current is required to start the motor of a car and hence cell cannot be used to start the motor.**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. (ρ _{Al} = 2.63 × 10^{−8} Ω m, ρ_{Cu} = 1.72 × 10^{−8} Ω m, Relative density of Al = 2.7, of Cu = 8.9.)**

**Solution :**Resistivity of aluminium, p_{1} = 2.63 × 10−8 Ω m

Relative density of aluminium, d_{1} = 2.7

Let l1 be the length of aluminium wire and m1 be its mass.

Resistance of the aluminium wire = R_{1}

Area of cross-section of the aluminium wire = A_{1}

Resistivity of copper, p_{2} = 1.72 × 10−8 Ω metre

Relative density of copper, d_{2} = 8.9

Let l2 be the length of copper wire and m2 be its mass.

Resistance of the copper wire = R2

Area of cross-section of the copper wire = A2

We can write,

R_{1} = (p_{1} × l_{1})/A_{1}

R_{2} = (p_{2} × l_{2})/A_{2}

Since R_{1} = R_{2}

(p_{1} × l_{1})/A_{1} = (p_{2} × l_{2})/A_{2}

And l_{1} = l_{2}

Therefore p_{1}/A_{1} = p_{2}/A_{2}

A_{1}/A_{2} = (2.63 × 10-8)/(1.72 × 10-8)

⇒ A_{1}/A_{2} = 2.63/1.72

Mass of the aluminium wire,

m_{1} = Volume × Density

= A_{1} × l_{1} × d_{1} = A_{1} l_{1} d_{1} …..1

Mass of the copper wire, m_{2} = Volume × Density

= A_{2} × l_{2} × d_{2} = A_{2} l_{2} d_{2} …….2

Dividing equation (1) by equation (2), we obtain

m_{1}/m_{2} = (A_{1} l_{1} d_{1})/(A_{2} l_{2} d_{2})

Since l_{1} = l_{2}

m_{1}/m_{2} = A_{1} d_{1}/A_{2} d_{2}

Since A_{1}/A_{2} = 2.63/1.72 calculated above

⇒ m_{1}/m_{2} = (2.63 × 2.7)/(1.72 × 8.9)

⇒ m1/m2 = 0.46

Since m1 is smaller than m2 Hence, aluminium is lighter than copper. Since aluminium is lighter, it is preferred for overhead power cables over copper.

**Question****17. What conclusion can you draw from the following observations on a resistor made of alloy manganin?**

CurrentA | VoltageV | CurrentA | VoltageV |

0.2 | 3.94 | 3.0 | 59.2 |

0.4 | 7.87 | 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 |

**Solution :**

It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistance of manganin is 19.7 Ω.

**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?**

**Solution :**

(a) When a steady current flows in a metallic conductor of non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant.

(b) No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.

(c) According to Ω ’s law V = I×R

If Voltage V is low, then Resistance R must be very low so that high current can be drawn from source.

(d) In order to prohibit the current from exceeding the safety limit, a high tension supply must have a very large internal resistance. If the internal resistance is not large, then the current drawn can exceed the safety limits in case of a short circuit.

**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 (10 ^{22}/10^{3}).**

**Solution :**

(a) Alloys of metals usually have greater resistivity than that of their constituent metals.

(b) Alloys usually have lower temperature coefficients of resistance than pure metals.

(c) The resistivity of the alloy, manganin, is nearly independent of increase of temperature.

(d) The resistivity of a typical insulator is greater than that of a metal by a factor of the order of 10^{22}.

**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) (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.**

**Solution :**

(a) Total number of resistors = n

Resistance of each resistor = R

**Question****21. Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1 Ω resistance.**

**Solution :**The resistance of each resistor connected in the given circuit, R = 1 Ω

Equivalent resistance of the given circuit = R’

The network is infinite. Hence, equivalent resistance is given by the relation,

R’ = 2 + (R’/R’ + 1)

or R’^{2} -2R’ -2 = 0

On solving R’ = (2+√4+8)/2

(Negative value of R’ cannot be accepted)

R’ = 1 + √3 Ω

R’ = 1 + 1.73 = 2.73 Ω

Internal resistance of the circuit, r = 0.5 Ω

Hence, total resistance of the given circuit = 2.73 + 0.5 = 3.23 Ω

Supply voltage, V = 12 V

According to Ω ’s Law, current drawn from the source is given by the ratio, = 12/3.23 = 3.72 A

**Question****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 up to 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) 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 1.0 V instead of 2.0 V?**

**(f ) 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?**

**Solution :**a) Constant emf of the given standard cell, E_{1} = 1.02 V

Balance point on the wire, l_{1} = 67.3 cm

A cell of unknown emf, ε, replaced the standard cell. Therefore, new balance point on the wire,

l = 82.3 cm

The relation connecting emf and balance point is,

E_{1}/l_{1} = E/l

E = l × E_{1}/l_{1}

E = 82.3 cm × 1.02 V/67.3 cm

E = 1.247 Volt.

The value of unknown emf is 1.247 V.

(b) Explanation:

The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.

(c) The balance point is not affected by the presence of high resistance.

(d) The point is not affected by the internal resistance of the driver cell.

(e) The method would not work if the driver cell of the potentiometer has an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then we cannot obtain balance point on the wire.

(f) The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a high percentage of error. The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.**Question****23. Figure 34 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?**

**Solution :**Resistance of the standard resistor, R = 10.0 Ω

Balance point for this resistance, l1 = 58.3 cm

Current in the potentiometer wire = i

Hence, potential drop across R, E_{1} = iR

Resistance of the unknown resistor = X

Balance point for this resistor, l_{2 }= 68.5 cm

Hence, potential drop across X, E_{2} = iX

The relation connecting emf and balance point is,

E_{1}/E_{2} = l_{1}/l_{2}

iR/iX = l_{1}/l_{2}

X = l_{1} × R/l_{2}

X = 68.5 × 10/58.3

X = 11.749 Ω

Therefore, the value of the unknown resistance, X, is 11.75 Ω .

Explanation:

If we fail to find a balance point with the given cell of emf, E, then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is lesser than the potential drop across the potentiometer wire AB, a balance point is obtained.