JEE Advanced (Single Correct MCQs): Current Electricity Chapter MCQs

KEY CONCEPT : The heat produced is given by

Thus heat (H) is doubled if both length (l) and radius (r) are doubled.

Since R_AB = 2R_CD therefore, current in AB will be half as compared to current in CD.

Here P₄ = Power dissipation in 4Ω

P₅ = Power dissipation in 5Ω

BC and AC are in series

Now BA and DC are in parallel.

Copper is a metal whereas Germanium is Semi-conductor.
NOTE : Resistance of metal decreases and semiconductor increases with decrease in temperature.

The equivalent circuits are shown in the figure.

The circuit represents balanced Wheatstone Bridge.
Hence 6R Ω resistance  is ineffective

For Max. Power External Resistance = Internal Resistance 2R = 4 Ω ∴ R = 2Ω

Since the opening or closing the switch does not affect the current through G, it means that in both the cases there is no current passing through S. Thus potential at A is equal to potential at B and it is the case of balanced wheatstone bridge..

I_P = I_Q and I_R = I_G

There will be no current flowing in branch BE in steady condition.
Let I be the current flowing in the loop ABCDEFA.
Applying Kirchoff's law in the loop moving in anticlockwise direction starting from C. + 2V – I (2R) – I (R) – V = 0
∴V = 3IR
⇒ I = V/3R ... (1)

Applying Kirchoff's law in the circuit ABEFA we get on moving  in anticlockwise direction starting from B

+ V + Vcap – IR – V = 0 (where Vcap is the p.d. across capacitor).

Let R be the resistance of wire.

Energy released in t second

Let R' be the resistance of the second wire

Dividing (i) by (ii)

or,        N ² = 18 × 2    ∴ N = 6

Since current I is independent of R6, it follows that the resistance R₁, R₂, R₃ and R₄ must form the balanced Wheatstone bridge.∴ R₁ R₄ = R₂R₃

The circuit is symmetrical about the axis POQ.
The circuit above the axis POQ represents balanced wheatstone bride. Hence the central resistance 2R is ineffective. Similarly in the lower part (below the axis POQ) the central resistance 2R is ineffective.

Therefore the equivalent circuit is drawn.

In ohm's law, we check V = IR where I is the current flowing through a resistor and V is the potential difference across that resistor. Only option (a) fits the above criteria.
NOTE : Remember that ammeter is connected in series with resistance and voltmeter parallel with the net resistance.

At null point

If radius of the wire is doubled, then the resistance of AC will change and also the resistance of CB will change. But since R₁/R₂ does not change so, R₃/R₄
 should also not change at null point. Therefore the point C does not change.

III < II < IV < I.

Total external resistance will be the total resistance of whole length of box. It should be connected between A and D.

For var ious combination s equivalent r esistan ce is maximum between P and Q.

KEY CONCEPT : The current in RC circuit is given by I = I₀e^–t/RC

On comparing with y = mx + C

When R is changed to 2R then slope increases and current becomes less. New graph is Q.

The current in 2W resistor will be zero because it is not a part of any closed loop.

KEY CONCEPT : At any instant of time t during charging process, the transient current in the circuit is given by

∴ Potential difference across resistor R is

∴ Potential diff. across C

KEY CONCEPT :

I_gG = (I – I_g) S

The heat supplied under these conditions is the change in internal energy Q = ΔU
The heat supplied Q = i²RT = 1 × 1 × 100 × 5 × 60 = 30,000 J = 30 kJ

Given X is greater than 2Ω when the bridge is balanced

When the resistances are interchanged the jockey shifts 20 cm. Therefore

The total charge enclosed in the dotted portion when the switch S is open is zero. When the switch is closed and steady state is reached, the current I coming from the battery is 9 = I (3 + 6) ⇒ I = 1A

∴ Potential difference across 3Ω resistance = 3V and potential difference across 6Ω resistance = 6V

∴ p.d. across 3 µF capacitor = 3V an d p.d. across 6 µF capacitor = 6V

∴ Charge on 3 µF capacitor Q1 = 3 × 3 = 9 µC Charge on 6 µF capacitor Q2 = 6 × 6 = 36 µC The total charge enclosed in the dotted portion =

∴ Charge passing the switch = 36 – 9 = 27 µC

We know that 
P=We know that 
P=V²/R
For constant value of potential difference (V) we have
For constant value of potential difference (V) we have

This is a case of balanced Wheatstone bridge R1 = 1Ω Case (ii)

Clearly the equivalent resistance (R2) will be less than 1Ω.
Case (iii)

Since, R₂ < R₁ < R₃

∴ P₂ > P₁ > P₃

We know that P=V²/R

For a given potential difference at a particular temperature

It is given that the powers of the bulbs are in the order 

100W > 60 W > 40W

c) The following points should be considered while making the circuit :

(i) An ammeter is made by connecting a low resistance R₂ in parallel with the galvanometer G₂.

(ii) A voltmeter is made by connecting a high resistance R₁ in series with the galvanometer G₁.

(iii) Voltmeter is connected in parallel with the test resistor R_T.

iv) Ammeter is connected in series with the test resistor RT.
(v) A variable voltage source V is connected in series with the test resistor R_T.

We know that R=ρ l/a

Where l is the length of the conductor through which the current flows and a is the area of cross section.

Here l = L and a = L × t

∴ R is independent of L