A resistor is connected to a battery of 12 ~V emf and internal resistance 2Omega. If the current in the circuit is 0.6…

Year	2024	Total
Questions	4	4

Year

2024

Total

Questions

Solution & Explanation

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 $i = 0.6 ~A$ $i = 0.6 \mathrm{~A}$

Step 1: Calculate Terminal Voltage

$V = E - ir$ $V = E - ir$

V = 12 - (0.6 \times 2)

V = 12 - 1.2 = 10.8 ~V

Pattern Recognition

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

Circuit with resistor and internal resistance for Q26

More Current Electricity Previous-Year Questions

Q9 neet_2026_03_may_morning Wheatstone Bridge

A uniform metallic wire having resistance

4Ω

is bent to form a square loop (ABCD) (see figure). A resistance of

2Ω

is connected between points B and D and a battery of

2 ~V

is connected across points A and C as shown in the figure. Now the value of current (

I

) is

Square loop circuit diagram with internal resistance for Q9 - NEET 2026 Code 12 — A 4 ohm wire bent into a square with an internal diagonal 2 ohm resistor and 2V battery.

A. $2 ~A$ $2 \mathrm{~A}$
B. $1.2 ~A$ $1.2 \mathrm{~A}$
C. $3 ~A$ $3 \mathrm{~A}$
D. $4.5 ~A$ $4.5 \mathrm{~A}$

Solution

Related Formula

Wheatstone bridge balance: (P)/(Q) = (R)/(S)

I = EReq

Core Logic

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 $1Ω$ $1\Omega$.

Deconstructed Wheatstone Bridge schematic Q9 — 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:

RABRBC = (1)/(1) = 1 and RADRDC = (1)/(1) = 1

Step 1: Simplify Equivalent Resistance

Because the ratios are equal, it forms a balanced Wheatstone bridge.

No current will flow through the central

2Ω

resistance connected between B and D.

The effective resistance across AC: Top branch: $Rₜₒₚ = 1Ω + 1Ω = 2Ω$ Bottom branch: $Rbottom = 1Ω + 1Ω = 2Ω$

Req = (2 \times 2)/(2 + 2) = 1 Ω

Step 2: Calculate Current

Using Ohm's Law:

I = EReq = 2 ~V1 Ω = 2 ~A

Pattern Recognition

Any symmetric square loop driven diagonally with a cross resistor constitutes a balanced Wheatstone bridge. Ignore the diagonal resistor completely.

Chapter Mix

Class 12 Physics: Current Electricity

Q16 neet_2026_03_may_morning Meter Bridge

In a metre bridge experiment (see figure), the positions of the cell, E, and galvanometer, G, are interchanged. We shall observe in the galvanometer:

Meter bridge circuit diagram with cell and galvanometer for Q16 — Standard meter bridge circuit indicating positions of E and G.

A. $Only the right-sided deflection$
B. $Only the left-sided deflection$ $\text{Only the left-sided deflection}$
C. $There will be no deflection irrespective of the position of the jockey$
D. $Both right-sided and left-sided deflection and at balance point, no deflection$

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.

Interchanged E and G in meter bridge showing identical balance state — 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

Q29 neet_2026_03_may_morning Electrical Energy and Power

A room heater is rated

400 ~W, 220 ~V

. If the supply voltage drops to

200 ~V

, what will be the power consumed (approximately)?

A. $121 ~W$ $121 \mathrm{~W}$
B. $331 ~W$ $331 \mathrm{~W}$
C. $200 ~W$ $200 \mathrm{~W}$
D. $400 ~W$ $400 \mathrm{~W}$

Solution

Related Formula

P = (V²)/(R)

PconsumedPrated = ( VappliedVrated )²

Core Logic

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 $Pc = (V²)/(R)$ $P_c = \frac{V^2}{R}$

Step 1: Calculate Consumed Power

Dividing the equations:

Pc = [ (V)/(V₀) ]² P₀

Pc = [ (200)/(220) ]² \times 400

Pc = ( (10)/(11) )² \times 400

Pc = (100)/(121) \times 400 = (40000)/(121) \approx 330.57 ~W

Step 2: Final Conclusion

The power consumed is approximately $331 ~W$ $331 \mathrm{~W}$.

Pattern Recognition

Power varies as the square of the voltage when resistance is fixed. If voltage drops by a small factor, power drops by the square of that factor.

Chapter Mix

Class 12 Physics: Current Electricity

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:

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Terminal Voltage

Pattern Recognition

Chapter Mix

Reference Study Guides

More Current Electricity Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Simplify Equivalent Resistance

Step 2: Calculate Current

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Consumed Power

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

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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:

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Terminal Voltage

Pattern Recognition

Chapter Mix

Reference Study Guides

More Current Electricity Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Simplify Equivalent Resistance

Step 2: Calculate Current

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Consumed Power

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

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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: