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4 months ago

Physics Current Electricity Class 12th

Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then T? is connected to T?. As the current in R = 6Ω attains a maximum value of steady state level, T? is disconnected from T? and immediately connected to T?. Potential drop across r = 3Ω resistor immediately after T? is connected to T? is _______ V. (Round off to the Nearest Integer)

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Vishal Baghel

Contributor-Level 10

When T? & T? are connected, at steady state current, I becomes
I = 6V/6Ω = 1A
Now when T? & T? are connected, the current through the inductor, just after connecting, remains same. So
V? Ω = 1A * 3Ω = 3volt.

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4 months ago

Physics Current Electricity Class 12th

An electric bulb rated as 200 W at 100 V is used in a circuit having 200 V supply. The resistance R that must be put in series with the bulb so that the bulb delivers the same power is _______ Ω.

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Raj Pandey

Contributor-Level 9

Given: Power in R_B = 200 W, R_B = 50 Ω, V_source = 200 V
I² R_B = 200
I² (50) = 200
I² = 4 ⇒ I = 2A
Also, I = V_source / (R + R_B)
2 = 200 / (R + 50)
2 (R + 50) = 200
R + 50 = 100
R = 50Ω

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4 months ago

Physics Current Electricity Class 12th

In the given figure, a battery of emf E is connected across a conductor PQ of length 'l' and different area of cross- sections by having radii r? and r? (r? < r?). Choose the correct option as one moves from P to Q:

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Vishal Baghel

Contributor-Level 10

i = constant
v ∝ E & E ∝ 1/r² So, E increases with decrease in the radius.
also v ∝ E
So, drift speed increases.

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4 months ago

Physics Current Electricity Class 12th

A 16Ω wire is bend to form a square loop. A 9V supply having internal resistance of 1Ω is connected across one of its sides. The potential drop across the diagonals of the square loop is __________ x 10?¹ V.

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alok kumar singh

Contributor-Level 10

Each side of the square has a resistance of 16/4 = 4Ω.
The side AC has the 9V, 1Ω source.
The other three sides (AB, BD, DC) form a path with resistance 4+4+4 = 12Ω.
This is in parallel with the side AC (4Ω).
Total resistance of the loop part: (12*4)/ (12+4) = 48/16 = 3Ω.
Total resistance of the circuit: R_total = 3Ω + 1Ω (internal) = 4Ω.
Total current from source I = V/R_total = 9/4 A.
This current splits.
Current through path ABDC, I? = I * (R_AC / (R_ABDC + R_AC) = (9/4) * (4/16) = 9/16 A.
Potential at B: V_B = V_A - I? R_AB = 9 - (9/16)*4 = 9 - 9/4 = 27/4 V.
Potential at D: V_D = V_C + I? R_CD = 0 + (9/16)*4 = 9/4 V.
Potential drop a

...more

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4 months ago

Physics Current Electricity Class 12th

The given potentiometer has its wire of resistance 10Ω. When the sliding contact is in the middle of the potentiometer wire, the potential drop across 2Ω resistor is :

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alok kumar singh

Contributor-Level 10

Applying nodal analysis at point X with potential xV:
(x-20)/5 + (x-0)/2 + (x-20)/5 = 0
2 (x-20) + 5x + 2 (x-20) = 0
9x - 80 = 0 => x = 80/9 V
Potential drop across 2Ω resistor = x = 80/9 V.
(Note: There seems to be a discrepancy in the provided solution and standard circuit analysis. The provided solution calculates the current junction, not a single point.)
The solution provided in the image calculates as:
(x-0)/5 + (x-20)/5 + (x-20)/2 = 0
2x + 2 (x-20) + 5 (x-20) = 0
9x - 140 = 0
=> x = 140/9 V
Potential drop across 2Ω = 20 - x = 20 - 140/9 = 40/9 V

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New answer posted

4 months ago

Physics Current Electricity Class 12th

In the given figure, there is a circuit of potentiometer of length AB = 10m. The resistance per unit length is 0.1Ω per cm. Across AB, a battery of emf E and internal resistance 'r' is connected. The maximum value of emf measured by this potentiometer is :
(The diagram shows a primary circuit with a 6V battery and a 20Ω resistor connected across the potentiometer wire AB. The secondary circuit shows an emf source being measured.)

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Raj Pandey

Contributor-Level 9

Maximum value of emf is measured when pointer is at B with I? = 0.
So, I_AB = 6 / (20 + 0.1 * 1000) = 6/120 = 1/20 A
V_AB = E = I_AB * R_AB
= (1/20) * 100
= 5V

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4 months ago

Physics Current Electricity Class 12th

In the given potentiometer circuit arrangement, the balancing length AC is measured to be 250 cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes 400 cm. The ratio of the emf of two cells, ε₁/ε₂,, is:

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alok kumar singh

Contributor-Level 10

ε? = 250 * Potential Gradient
ε? + ε? = 400 * Potential Gradient
(ε? )/ (ε? + ε? ) = 250/400 = 5/8
By solving above
ε? /ε? = 5/3

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4 months ago

Physics Current Electricity Class 12th

A current through a wire-depends on time as

i = a₀t + bt²

Where a₀ = 20A/s and b = 8As^-2. Find the charge crossed through a section of the wire in 15s.

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alok kumar singh

Contributor-Level 10

$\frac{d q}{d t} = t = 20 t + 8 t^{2}$

$\Rightarrow \int_{0}^{q} d q = \int_{0}^{15} (20 t + 8 t^{2}) d t$

$\Rightarrow q = 20 * \frac{1 5^{2}}{2} + 8 . \frac{1 5^{3}}{3} = 11250 C$

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4 months ago

Physics Current Electricity Class 12th

A cell E₁ of emf 6V and internal resistance $2 Ω$ is connected with another cell E₂ of emf 4V and internal resistance $8 Ω$ (as shown in the figure). The potential difference across points X and Y is: