Test: MCQs (One or More Correct Option): Current Electricity | JEE Advanced - JEE MCQ

KEY CONCEPT : During decay of charge in R.C. circuit, I = I₀e^–t/RC

Since potential difference between the plates is same initially therefore I is same in both the cases at t = 0 and is equal to

or,      t₁ = 0.5 t₂

Therefore time taken for the first capacitor (1 µF) for discharging 50% of initial charge will be less.

NOTE : The conductivity of a semiconductor increases with increase in temperature i.e. the resistivity decreases with increase in temperature.

In a conducting solid, the collisions become more frequent with increase of temperature.

The net resistance of the circuit is 9W as shown in the following figures.

The flow of current in the circuit is as follows.

NOTE : The current divides into two equal parts if passes through two equal resistances in parallel.
Thus current through 4Ω resistor is 0.25 A.

At 0K an insulator does not permit any current to flow through it. Option (a) is correct.
At 0K a semiconductor behaves as an insulator. Option (b) is correct.
In reverse biasing at 300 K, a very small current. flows through a p-n junction diode. Option (d) is correct.

In case of metal, the current flowing will be very-very high because a metal becomes super conductor at 0K. Option (c) is incorrect

⇒ option (a) is correct.
The potential difference accros R_L= potential difference accros R_p

_⇒ option (b) is incorrect.

_⇒ option (c) is incorrect.
The magnitude of power dissipated accross R₂ is

Now when R₁ and R₂ are interchanged the equivalent resistance between R₁ and R_L

∴ Potential drop across this equivalent resistance

∴ Potential difference accros 

∴ The magnitude of the power dissipation in R_L will decrease by a factor 9 if R₁ and R₂ are interchanged. (d) is the correct option.

The given circuit is an extension of wheatstone bridge, therefore points P and Q are at the same potential and point S and T are also at the same potential. Therefore no current passes through PQ and ST and the circuit reduces to as shown

   = 3A

As P and Q are equipotential and potential at S is less than the potential at P (potential drops across a resistance as current passes through it), therefore V_S < V_Q.

When resistances are connected in series

From (i) and (ii) t₂ = 2 min. Therefore (b) is correct.
When resistance are connected in parallel

From (i) and (iii) t₂ = 0.5 min

∴ (d) is correct

Applying KVL in MNOPQAM

V₁ - iR₁ + V₂ - iR₃=0

Applying KVL in NOPQN

If V₁ = V₂ then R₁ = R₃ = R₂∴ (a) is correct option.
If V₁ = V₂ then R₁ = R₃ = 2R₂∴ (b) is correct option. (R₂ can have any value as there is no current flowing through d) If V₁ = 2V₂ then 2R₃ = R₁ ∴ (c) is incorrect option.
If 2V₁ = V₂ then R₃ = 2R₁ = R, ∴ (d) is correct option.

With the use of filament and the evaporation involved, the filament will become thinner thereby decreasing the area of cross-section and increasing the resistance. Therefore the filament will consume less power towards the end of life.
As the evaporation is non-uniform, the area of cross-section will be different at different cross-section. Therefore temperature distribution will be non-uniform. The filament will break at the point where the temperature is maximum.
When the filament temperature is higher  it emits light of lower wavelength or higher band of frequencies.

At t=0, Capacitors act as short circuit and voltmeter display –5v At t = µ, Capacitor acts as open circuit and no current flows through voltmeter ( ∴ very high resistance of voltmeter) so it display +5V. (a) is the correct option

(b) is the correct option

∴ (c) is the correct option After  a long time no current flows since both capcitor and voltmeter do not allow current to flow.

∴ (d) is the correct option.