Physics Current Electricity
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New answer posted
a month agoContributor-Level 10
Heat generated in the resistance
H = i²RT
H? = 500 = (1.5)²R (20) ⇒ R = 500/45 = 11.11 Ω
H? = (3)²R (20) = 9 * (500/45) * 20 = 2000 J.
New answer posted
a month agoContributor-Level 10
R = R? [1 + α? T]
⇒ 16 = R? [1 + α (15 - T? )]
20 = R? [1 + α (100 - T? )]
Assuming T? = 0°C:
16 = R? (1 + 15α)
20 = R? (1 + 100α)
Dividing the equations:
16/20 = (1+15α)/ (1+100α)
16 + 1600α = 20 + 300α
1300α = 4
α = 4/1300 ≈ 0.003 °C? ¹
New answer posted
a month ago
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.
New answer posted
a month agoContributor-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
New answer posted
a month ago
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.
New answer posted
a month agoContributor-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
New answer posted
a month ago
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
New answer posted
a month ago
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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