Overview

Get insights from 140 questions on Overview, answered by students, alumni, and experts. You may also ask and answer any question you like about Overview

Follow Ask Question
140

Questions

0

Discussions

4

Active Users

0

Followers

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

d q d t = t = 2 0 t + 8 t 2

0 q d q = 0 1 5 ( 2 0 t + 8 t 2 ) d t

q = 2 0 * 1 5 2 2 + 8 . 1 5 3 3 = 1 1 2 5 0 C

 

New answer posted

a month ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

i = 6 4 2 + 8 = 0 . 2 A

V x + 4 + 0 . 2 * 8 = V Y

V Y V X = 5 . 6 V

 

New answer posted

a month ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Potential difference across 2k Ω  is 5V, thus current through it,

  i = 5 2 * 1 0 3 = 2 5 * 1 0 4 A .          

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

R i = ρ l A

R f = ρ ( 1 . 2 5 l ) ( A / 1 . 2 5 ) = ( 1 . 2 5 ) 2 * ρ l A

R f = 1 . 5 6 2 5 * R i

R f R l R l * 1 0 0 = 5 6 . 2 5 %

New answer posted

a month ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

W = q Δ V = 2 0 * 1 5 = 3 0 0 J

New answer posted

a month ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

Given circuit can be re-drawn and it becomes case of balanced Wheatstone bride

R A B = ( 2 R ) ( 2 R ) 2 R + 2 R = R

New answer posted

a month ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

radius r = 0.5 mm

σ = 5 * 1 0 7 s / m

E = 1 0 * 1 0 3 V / m = 1 0 2 V / m

As we know

J = σ E

J = i A = σ E

i = σ E A = 5 * 1 0 7 * 1 0 2 * π * ( 0 . 5 * 1 0 3 ) 2 = 1 . 2 5 π * 1 0 1 A = 1 2 5 π m A = x 3 π m A

 x = 5

New answer posted

a month ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

E 1 E 2 = 3 8 0 7 6 0 = 1 2                                                                                                      

New answer posted

a month ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

i2 = 2i1

i1 + i2 = 6

i1 = 2A.

New answer posted

a month ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Power Consumed = P = V 2 R

P A P B = R B R A

R A = 2 R B

For Series Combination

P S = V 2 3 R B

For Parallel Combination

P P = 3 V 2 2 R B

P S P P = 2 9

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.