Test: Current Electricity - 1 - NEET MCQ

Kirchhoff’s loop rule is based on the principle of conservation of energy. The work done in transporting a charge in a closed loop is zero. The algebraic sum ( since potential differences can be both positive and negative) of potential differences around any closed loop is always zero.

The balance point of the Wheatstone’s bridge is determined by the ratio of the resistances. The change in the emf of the external battery will have no effect on the balance point.
 

Explanation:

• Initial Balanced Wheatstone Bridge: In the initial balanced Wheatstone bridge configuration, the emf of the cell is 1.46 V and all four resistors R1, R2, R3, and R4 are set to specific values to achieve balance.


• Replacement of Cell: When the cell is replaced by another cell with an emf of 1.08 V, the balance of the Wheatstone bridge is disrupted.


• Requirement for Rebalancing: In order to rebalance the Wheatstone bridge with the new cell of emf 1.08 V, no resistance needs to be changed.


• Reasoning: The balance of the Wheatstone bridge is determined by the ratio of the resistances in the bridge arms and not by the absolute values of the resistances. As long as the ratio of the resistances remains the same, the balance will be maintained regardless of the emf of the cell.


• Conclusion: Therefore, in this scenario, no resistance needs to be changed to obtain the balance again with the new cell of emf 1.08 V.


 

We want the Voltage difference V_AB.
let A and B be open and not connected to any thing.
There is a current that flows from cell of larger emf to the cell of small emf.
Call that current as  I Amperes.

1.25 V - 0.75 V - I * R2 - I * R1 = 0

I = 0.5 / (R1+R2)
V_AB =  -0.75 - I * R2 =  - 0.75 - 0.5 * R2 / (R1+R2)

= - (0.75 R1 + 1.25 R2) / (R1+R2)

= - ().75 * 1 + 1.25 * 1) / (1 + 1)  volts

=  - 1 volts

Power P= V²/R​
If length is reduced by 10% then, new resistance of filament will be R′.
R′ = R−10% of R
R′ = 0.9R
Now new power of heater is P_2​
P_2​ = V²​​/R′ = V²/0.9R ​ ​= 1.1 P
% increase powe r = 11%

Total current through the wire is given by:

I = ∫₀^R J(r)·2πr dr

Substitute the given current density:

I = ∫₀^R J₀ (1 − r²/R²) · 2πr dr

Take constants outside the integral:

I = 2πJ₀ ∫₀^R (1 − r²/R²) r dr

Split the integral:

I = 2πJ₀ [ ∫₀^R r dr − (1/R²) ∫₀^R r³ dr ]

Evaluate the integrals:

∫₀^R r dr = R²/2,    ∫₀^R r³ dr = R⁴/4

Substitute the values:

I = 2πJ₀ [ R²/2 − R²/4 ] = 2πJ₀ · R²/4 = πJ₀R²/2

Final Answer: I = πJ₀R²/2

Voltage across 5Ω=10 V

∴ I = 10/5 A = 2 A

The electrons in a conductor have random velocities and when an electric field is applied, they suffer repeated collisions and in the process move with a small average velocity, opposite to the direction of the field. This is equivalent to positive charge flowing in the direction of the field.

Potentiometer measures the potential difference using null deflection method, where no current is drawn from the cell; whereas voltmeter needs a small current to show deflection. So, accurate measurement of p.d is done using a potentiometer.

• Copper is a conductor and we know that for conductors, resistance is directly proprtional to temperature. Therefore on decreasing temperature resistance also decreases.
• Whereas, germanium is a semiconductor and for semiconductors, resistance is inversely proportional to temperature.
• So on decreasing temperature resistance increases.

The network is a Wheatstone bridge. It’s balanced because

​,

so the potential at the top and bottom junctions is the same. Hence no current flows through the middle resistor, whatever its value (10 Ω or 20 Ω).

So the equivalent seen by the battery is just the two series paths in parallel:

With the original current 4 A, the battery voltage is

Replacing 10 Ω with 20 Ω doesn’t change R_eq​, so the current drawn remains

If the current due to the auxillary battery is decreased, the potential gradient will be decreases, so the balancing length increases.Thus null point will move to fifth wire.

We know that R=V/I_g−G
The voltmeter gives the full-scale deflection for potential difference V. Its resistance is G. Hence, I_n=(V/G).
Given that V=nV

With a known resistance in one of the gaps, the meter bridge is used to determine the value of unknown resistance by the formula. 

Resistors 20Ω,100Ω and 25Ω will be in parallel. Their equivalent is 10Ω.


p.d. across 10Ω,10I = 10 × 10 = 100 V
This will be the voltmeter reading. Also, this will be the p.d. across each of 20Ω,100Ω and 25Ω resistors.
Ammeter reading = current through 25Ω=100/25=4 A.

So, 400 = R₀[1 + 0.005 × 2000]

• A potentiometer is considered to be sensitive if the potential gradient dV/dl is low.
• Such a potentiometer can measure very small changes in potential difference.
• Increasing the length of the potentiometer wire decreases the potential gradient. Its sensitivity increases. Increasing potential gradient decreases the sensitivity.
• Increasing the emf of the primary cell and by decreasing the length, potential gradient increases.

Suppose V is the voltage of the supply and R is the resistance of each bulb.
Now, R_p = R/3 and current in ammeter, I = V/R_p = 3 V/R, provided all three bulbs are working properly.
If one bulb has broken down, then R_p = R/2 and I = 2V/R
Therefore, current decreases and since current through each bulb is V/R the same as before, brightness of bulbs is not affected.

60/40 = l/100-l, = R/8

so R=12

now changed position 8 is proportional to l and 12 to (100-l)

8/12 = l/100-l

l = 40ohm