Class 12th
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New answer posted
10 months agoContributor-Level 10
L = 0.07H and R = 12Ω
Vs = 220V, 50Hz
XL = 2πfL = 2 (22/7) (50) (0.07) = 22Ω
Z = √ (R² + XL²) = √ (12² + 22²) = √ (144+484) = √628 ≈ 25Ω
i = V/Z = 220/25 = 8.8A
tan φ = XL/R = 22/12 = 11/6
φ = tan? ¹ (11/6)
New answer posted
10 months 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
10 months agoContributor-Level 9
I = I? e? /?
= (20/10000) e^- (1*10? / 10*10? ³)
= 2 * 10? ³ e? ¹
The provided solution calculates as:
= 2 * 10? ³ e? ¹
= 2e? ¹ mA
= 2 * 0.37mA
= 0.74 mA = 74/100 mA
(Note: There seems to be a calculation discrepancy in the source image, the steps shown lead to I = 2e? ¹ mA)
New answer posted
10 months agoContributor-Level 10
β = λD/d or β ∝ λ
Also, we know that λ_blue < _orange
So, β_orange > β_blue
So, dist b/w consecutive fringes will decrease.
New answer posted
10 months agoContributor-Level 9
F_net = -kq²/ (l+x)² + kq²/ (l-x)²
= kq² [ (l+x)² - (l-x)² / (l²-x²)² ]
= kq² [ 4lx / (l²-x²)² ]
For x << l,
ma ≈ -4kq²x / l³
a = - (4kq²/ml³)x
ω² = 4kq²/ml³
ω = √ (4kq²/ml³)
= √ (4 * 9*10? * 10 / (1*10? * 1)
= 6 * 10? rad/s
= 6000 * 10? rad/s

New answer posted
10 months agoContributor-Level 10
As charge remains same. So,
Initial total charge Q = (2C)V + CV = 3CV
When dielectric is inserted in C, its new capacitance is KC.
The capacitors 2C and KC are in parallel.
Equivalent capacitance C_eq = 2C + KC = (K+2)C
New potential Vc = Q/C_eq = 3CV/ (K+2)C) = 3V/ (K+2)
New answer posted
10 months agoContributor-Level 10
v = 1/√με = 1/√ (µ? µ? ε? ε? ) = c/√ (µ? ε? )
v = 3*10? / √ (1*81)
v = 3*10? / 9 m/s
= 0.33 * 10? m/s
= 3.33 * 10? m/s
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