NCERT Questions for Class 12 Physics Chapter 3 – Current Electricity

Important Questions with Solutions of Class 12 Physics Chapter 3 – Current Electricity

1) Resistivities of copper, silver & manganin are 1.7 x 10^-8 Ωm, 1.0 x 10^-8 Ωm & 44 x 10^-8 Ωm respectively. Find out which one of these is the best conductor.

Ans – Resistance is directly proportional to desired resistance (resistivity) when length l & area A are made constant i.e, R =ρl/A.

Therefore, silver is the best conductor since its resistance is low.

2) Two hot wires of the same size are first linked in series & then parallel to the supply source. Calculate the heat ratio in the 2 cases.

Ans – As you know,

H = I²Rt (I = V/R)

⇒ H ∝ 1/R

Thus, the heat output ratio is 1:4.

3) A series of n-identical resistors where every resistance is calculated RΩ when linked in series got an effective resistance of XΩ & once the resistors are merged in parallel they have an effective resistance of YΩ. Calculate between R, X & Y.

Ans – As you know,

n-resistors when linked in series => X=nR …(1)

n-resistors when connected in parallel => Y=R/n …(2)

Multiply both the equations (1) & (2),

=> XY = nR*R/n

=> XY = R²

4) The figure shows a section of a pure semiconductor S in series having a variable resistor R & a source of constant voltage V. Will you increase & decrease the value of R to read the same value in ammeter (A) during the heating of the semiconductor S? Justify your reasons!

Ans – The resistance of a semiconductor reduces when the temperature increases. Hence, the semiconductor S is heated to raise the temperature levels.

The circuit’s total resistance should not be varied for a constant current flow in the ammeter. Therefore, you must increase the value of R.

5) A battery of emf E & internal resistance r passes a current of I₁ & I₂ when linked with an external resistance of R₁ & R₂. Calculate the battery’s internal resistance & EMF.

Ans – As you know,

E is the emf of the battery, and r is the internal resistance which carries the currents I₁, I₂ & R₁, R₂ are the external resistances.

Current I₁ can be denoted as: I₁ = E/(R₁+r)

E = I₁ (R₁+r) …(1)

In the same way E = I₂ (R₂+r) …(2)

From the equation 1 & 2,

I₁ (R₁+r) = I₂ (R₂+r)

I₂r – I₁r = I₁R₁ – I₂R₂

r (I₂-I₁) = I₁R₁ – I₂R₂

r = (I₁R₁-I₂R₂)/(I₂-I₁)

Now the EMF of the battery

E = I₁(R₁+r)

So, the battery’s internal resistance & EMF are,

6) a) 3 resistors 1Ω, 2Ω & 3Ω are linked in series. Calculate the overall resistance of the connection.

Ans – From the given data, 3 resistors of resistances 1Ω, 2Ω & 3Ω are connected in series. The total resistance of the linked is the algebraic sum of the individual resistances. So, the total resistance is 1 + 2 + 3 = 6Ω.

b) If the values are connected to a 12V EMF battery & neutral internal resistance, find the power drop across every resistor.

Ans – Assume that the current flowing in the circuit is the I.

EMF of the battery E = 12V

The total resistance of the circuit R = 6Ω

Using Ohm’s law, the current relationship is given by: I = E/R

I = 12/6 = 2A

Imagine the potential drop across 1Ω the resistor as V₁

Using Ohm’s law, V₁ can have the obtained value V₁ = 2 x 1 = 2V …(1)

Consider the potential drop across 2Ω resistor as V₂

Using Ohm’s law, the value of V₂ can be calculated as V₂ = 2 x 2 = 4V …(2)

Consider the potential drop across a 3Ω resistor as V₃

Then with Ohm’s law, V₃ = 2 x 3 = 6V …(3)

So the potential drops across resistors 1Ω, 2Ω & 3Ω are 2V, 4V & 6V respectively.

7) A small current is passed through a 15 m-long wire with a cross-section of 6.0 x 10^-7 m² & its resistance is 5.0Ω. Calculate the material resistivity at a temp of this survey.

Ans – As we know from above,

The length of the wire is l = 15m

Cross-section area of the wire a = 6.0 x 10^-7 m²

Material resistance of the wire R = 5.0Ω

Let p be the resistance of the material in the wire

Since the resistance is linked with resistivity,

R = ρ l/A

ρ = RA/l

ρ = (5 x 6.0 x 10^-7)/15

ρ = 2 x 10^-7 m² So, the internal resistance of the material is 2 x 10^-7 m².

8) A nichrome-driven heating element connected to a 230 V supply draws an initial current of 3.2A which, when held up for a few seconds, has a constant current of 2.8A. If the room has an internal temperature of 27°C then calculate the constant temperature of the element. Given the nichrome’s temperature coefficient of resistance over the temperature range is denoted as 1.70 x 10^-1C^-1.

Ans – From the equation above, the supply voltage will be V = 230V

The initial current drawn is I₁ = 3.2A

Let R₁ be the initial resistance.

Using Ohm’s law, R₁=V/I₁

R₁ = 230/3.2 = 71.87Ω

Steady-State Value of Current I₂ = 2.8A

Let the steady-state resistance be R.

Using Ohm’s law, R₂ = V/I₂

R₂ = 230/2.8 = 82.14Ω

Let the internal temperature coefficient of nichrome be α = 1.70 x 10^-4 C^-1

Then the initial temperature of nichrome will be T1 = 27 C

Let the steady state heat of attainment of nichrome be T2

Now use the formula of α,

T₂ – 27 = 840.5

T₂ = 867.5 C

So, the static temperature of a heating element is 867.5°C.

9) Consider a silver with 2.1Ω resistance at 27.5°C & 2.7Ω resistance at 100°C. Calculate the thermal resistance coefficient of silver.

Ans – As you know,

Temperature T₁ = 27.5°C

The resistance of the silver wire at T₁ is R₁ = 2.1Ω

Temperature T₂ = 100°C

Resistance of silver wire at T₂ is R₂ = 2.7Ω

Assume the internal temperature coefficient of silver is α.

The relation between temperature & resistance is given by the formula,

Thus, the coefficient of heating of silver is 0.0039° C^-1.