A uniform metallic wire having resistance 4Ω$4\Omega$ is bent to form a square loop (ABCD) (see figure). A resistance of 2Ω$2\Omega$ is connected between points B and D and a battery of 2 ~V$2 \mathrm{~V}$ is connected across points A and C as shown in the figure. Now the value of current (I$I$) is
A 4 ohm wire bent into a square with an internal diagonal 2 ohm resistor and 2V battery.
The wire has a total resistance of 4Ω$4\Omega$ and is bent into a square ABCD. Thus, each side of the square has a resistance of 1Ω$1\Omega$.
A 4 ohm wire bent into a square with an internal diagonal 2 ohm resistor and 2V battery.
The battery is connected across nodes A and C. A 2Ω$2\Omega$ resistor is connected across nodes B and D.
We can unfold this circuit diagram:
Branches AB and BC form two series resistors on top (totaling 1Ω + 1Ω = 2Ω$1\Omega + 1\Omega = 2\Omega$, but tapped at B).
Branches AD and DC form two series resistors on bottom.
The ratio of the arms is:
Because the ratios are equal, it forms a balanced Wheatstone bridge.
A 4 ohm wire bent into a square with an internal diagonal 2 ohm resistor and 2V battery.
No current will flow through the central 2Ω$2\Omega$ resistance connected between B and D.
In a metre bridge experiment (see figure), the positions of the cell, E, and galvanometer, G, are interchanged. We shall observe in the galvanometer:
Standard meter bridge circuit indicating positions of E and G.
A.Only the right-sided deflection$\text{Only the right-sided deflection}$
B.Only the left-sided deflection$\text{Only the left-sided deflection}$
C.There will be no deflection irrespective of the position of the jockey$\text{There will be no deflection irrespective of the position of the jockey}$
D.Both right-sided and left-sided deflection and at balance point, no deflection$\text{Both right-sided and left-sided deflection and at balance point, no deflection}$
Solution
Core Logic
A meter bridge is a practical application of a Wheatstone bridge.
According to the properties of a Wheatstone bridge, if the positions of the battery (cell, E) and the galvanometer (G) are interchanged, the balance point remains unchanged.
Standard meter bridge circuit indicating positions of E and G.
Step 1: Final Conclusion
Because the new configuration still acts as a valid bridge circuit, the galvanometer will still show bidirectional deflection (left and right) depending on which side of the null point the jockey is pressed. At the precise balance point, it will show zero deflection.
Pattern Recognition
Interchanging battery and galvanometer in a Wheatstone bridge does not disturb the balance condition. It just re-maps the input-output nodes symmetrically.
Chapter Mix
Class 12 Physics: Current Electricity
Q26neet_2026_03_may_morningCells EMF and Internal Resistance
A resistor is connected to a battery of 12 ~V$12 \mathrm{~V}$ emf and internal resistance 2Ω$2\Omega$. If the current in the circuit is 0.6 ~A$0.6 \mathrm{~A}$, the terminal voltage of the battery is:
A.10 ~V$10 \mathrm{~V}$
B.10.8 ~V$10.8 \mathrm{~V}$
C.12 ~V$12 \mathrm{~V}$
D.1.2 ~V$1.2 \mathrm{~V}$
Solution
Related Formula
V = E - ir$V = E - ir$
Core Logic
When a battery discharges (supplies current to a resistor), its terminal voltage drops due to the potential lost across its own internal resistance.
Given:
EMF E = 12 ~V$E = 12 \mathrm{~V}$
Internal resistance r = 2 Ω$r = 2 \Omega$
Circuit current i = 0.6 ~A$i = 0.6 \mathrm{~A}$
Terminal voltage during discharge is strictly E - ir$E - ir$. Do not confuse this with charging, where terminal voltage is E + ir$E + ir$.
Chapter Mix
Class 12 Physics: Current Electricity
Q29neet_2026_03_may_morningElectrical Energy and Power
A room heater is rated 400 ~W, 220 ~V$400 \mathrm{~W}, 220 \mathrm{~V}$. If the supply voltage drops to 200 ~V$200 \mathrm{~V}$, what will be the power consumed (approximately)?
The resistance R$R$ of the heater remains constant regardless of the voltage applied.
Rated power P₀ = (V₀²)/(R)$P_0 = \frac{V_0^2}{R}$
Consumed power Pc = (V²)/(R)$P_c = \frac{V^2}{R}$
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
