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JEE Physics
Current Electricity

Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

13 Q 9 Topics Take Mock Test
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Topics in JEE Physics
In a Wheatstone bridge, arms P, Q, R, and S have resistances 10Ω, 15Ω, 20Ω, and 30Ω respectively. A galvanometer is connected between junctions of P-Q and R-S. The galvanometer reading will be:
A Zero (balanced condition)
B Non-zero, indicating unbalanced bridge
C Depends on EMF of battery
D Maximum deflection
Correct Answer:  A. Zero (balanced condition)
EXPLANATION

For balanced bridge: P/Q = R/S. Check: 10/15 = 20/30 → 2/3 = 2/3. The bridge is balanced, so no current flows through galvanometer (zero reading).

Test
A superconductor exhibits zero resistance below its critical temperature because:
A Electrons move without collision
B Free electrons pair up (Cooper pairs) forming a quantum state
C Thermal energy is insufficient for scattering
D Atoms are in rigid lattice
Correct Answer:  B. Free electrons pair up (Cooper pairs) forming a quantum state
EXPLANATION

BCS theory explains superconductivity: below critical temperature, electrons form Cooper pairs with no scattering, resulting in zero resistance.

Test
The equivalent resistance of an infinite ladder network of 1Ω resistors (each rung) is:
A
B
C (1+√5)/2 Ω
D Infinite
Correct Answer:  C. (1+√5)/2 Ω
EXPLANATION

For infinite ladder: Rₑq = 1 + (Rₑq×1)/(Rₑq+1). Solving: Rₑq² - Rₑq - 1 = 0, giving Rₑq = (1+√5)/2 ≈ 1.618Ω (Golden ratio)

Test
The length of a conductor increases by 10% when stretched. Assuming volume remains constant, the resistance increases by approximately:
A 10%
B 20%
C 21%
D 100%
Correct Answer:  C. 21%
EXPLANATION

R = ρL/A. If L increases by 10% and volume constant, A decreases by ~9.1%. New R = 1.1R₀/0.91 ≈ 1.21R₀, so 21% increase.

Test
When a superconductor is cooled below its critical temperature, its resistance becomes zero. What is the effect on current flowing through it when connected to a constant voltage source?
A Current becomes infinite
B Current becomes zero
C Current remains same
D Cannot be determined
Correct Answer:  A. Current becomes infinite
EXPLANATION

With R = 0, by I = V/R, current becomes infinite. In practice, voltage source cannot be maintained across a superconductor

Test
The electric field inside a copper conductor carrying current is approximately:
A Zero
B E = ρJ, where ρ is resistivity and J is current density
C Equal to the applied EMF
D Inversely proportional to conductivity
Correct Answer:  B. E = ρJ, where ρ is resistivity and J is current density
EXPLANATION

By Ohm's law in microscopic form: E = ρJ. Inside the conductor, electric field maintains the drift of electrons

Test
A battery of EMF 12V and internal resistance 2Ω is connected to a load resistance R. For maximum power transfer, R should be:
A
B
C 12Ω
D 24Ω
Correct Answer:  B. 2Ω
EXPLANATION

Maximum power transfer theorem: Load resistance equals internal resistance. R = r = 2Ω for maximum power

Test
Which of the following best explains why semiconductor resistance decreases with increase in temperature?
A Increase in collision frequency
B Decrease in thermal energy
C Increase in number of free charge carriers
D Increase in electron mass
Correct Answer:  C. Increase in number of free charge carriers
EXPLANATION

In semiconductors, increased temperature promotes more electrons from valence to conduction band, increasing carrier concentration and decreasing resistance

Test
A heating element of resistance R is connected to a battery of EMF E and internal resistance r. Maximum power is dissipated in R when:
A R = 0
B R = r
C R > r
D R approaches infinity
Correct Answer:  B. R = r
EXPLANATION

By maximum power transfer theorem, maximum power is transferred to external load when load resistance equals internal resistance

Test
The equivalent resistance between two opposite corners of a cube made of 12 wires of 1Ω each is:
A
B 5/6Ω
C 7/12Ω
D 2/3Ω
Correct Answer:  B. 5/6Ω
EXPLANATION

Due to symmetry, current divides into three paths of 1Ω each in parallel at the first vertex, then similar distribution at other vertices. Equivalent = 1/3 + 1/6 + 1/3 = 5/6Ω

Test
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