CBSE Board Level — 10 Questions
The unit of resistivity is:
A wire of resistance R is stretched to double its length (volume constant). Its new resistance is:
Terminal voltage of a cell equals its EMF when:
Three resistors of 2Ω, 3Ω, and 6Ω are connected in parallel. The equivalent resistance is:
In a potentiometer experiment, the balancing length for a cell with EMF 1.5V is 60 cm. What is the EMF of another cell that balances at 40 cm?
A cell of EMF 2V and internal resistance 0.5Ω is connected to a 3.5Ω resistor. Calculate the terminal voltage across the resistor.
I = ε/(R+r) = 2/(3.5+0.5) = 2/4 = 0.5 A
V = IR = 0.5 × 3.5 = 1.75 V
Or: V = ε − Ir = 2 − 0.5×0.5 = 2 − 0.25 = 1.75 V ✓
NEET Standard — 8 Questions
In a Wheatstone bridge, P = 10Ω, Q = 15Ω, R = 20Ω. For balance, S should be:
A 40W and 100W bulb are connected in series to 220V supply. Which bulb glows brighter?
Drift velocity of electrons in a conductor is of the order of:
5 cells each of EMF 2V and internal resistance 0.5Ω are connected in parallel to a 5Ω external resistance. Find the current through the external resistance.
Parallel cells: ε_eff = ε = 2V, r_eff = r/n = 0.5/5 = 0.1Ω
I = ε_eff/(R + r_eff) = 2/(5+0.1) = 2/5.1 ≈ 0.392 A
JEE Main — 6 Questions
In a circuit, two identical resistors R are connected with a battery of EMF ε and internal resistance r. When connected in series: I_s = ε/(2R+r). When in parallel: I_p = ε/(R/2+r). The ratio I_p/I_s equals:
In a meter bridge experiment, the null point is found at 40 cm. If the known resistance is 30Ω, the unknown resistance S = ? (in Ω)
S = R(100−l)/l = 30×(100−40)/40 = 30×60/40 = 30×1.5 = 45 Ω
In a circuit: battery 1 (10V, 1Ω), battery 2 (6V, 1Ω) connected in series with R = 3Ω. Both batteries oppose each other. Find the current.
Net EMF = 10 − 6 = 4V (opposing) | Total resistance = 1+1+3 = 5Ω
I = 4/5 = 0.8 A (in direction of larger EMF, 10V battery)
JEE Advanced — 4 Problems
Which of the following CORRECTLY describes a Wheatstone bridge at balance? (Select ALL that apply)
Correct: A, B, C
A ✓: Standard balance condition. B ✓: Null detector shows zero at balance. C ✓: Equal potential at B and D means no current between them. D ✗: At balance, galvanometer resistance does NOT affect the balance condition (only sensitivity).
A network has 5 resistors, each of resistance 1Ω, arranged as follows: R₁ from A to B, R₂ from B to C, R₃ from A to D, R₄ from D to C, R₅ from B to D. Find equivalent resistance from A to C (in Ω, rounded to 1 decimal).
This is a Wheatstone-like bridge. Apply KVL. Let V_A = 1V, V_C = 0. Using KCL at B and D:
At B: (V_A−V_B)/R₁ = (V_B−V_C)/R₂ + (V_B−V_D)/R₅
At D: (V_A−V_D)/R₃ + (V_B−V_D)/R₅ = (V_D−V_C)/R₄
By symmetry: not balanced (R₁=R₂=R₃=R₄=R₅=1). Solve: V_B = 3/5, V_D = 2/5
I_total = (V_A−V_B)/1 + (V_A−V_D)/1 = (1−3/5)+(1−2/5) = 2/5+3/5 = 1
R_eff = V_A/I = 1/1... wait, use I_total through A: I = 1A for V=1V → R = 0.7 Ω