Class 12th

Get insights from 11.9k questions on Class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 12th

Follow Ask Question
11.9k

Questions

0

Discussions

44

Active Users

0

Followers

New answer posted

7 months ago

0 Follower 17 Views

A
alok kumar singh

Contributor-Level 10

In steady state, capacitor branch is open.
V_AB = 2Ω (1A) + 2Ω (3A) = 8V.

 

New answer posted

7 months ago

0 Follower 19 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image 

 

New answer posted

7 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

V_PQ * (49/100). 1.02 = V_PQ * 0.49. V_PQ ≈ 2.08V.
Potential gradient = 2.08V/100cm = 0.0208 V/cm.

New answer posted

7 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image 

 

New answer posted

7 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

Propagation direction v? = E? * B? .
v? = k? * (2i-2j) = 2 (k*i) - 2 (k*j) = 2j + 2i.
Unit vector is (i+j)/√2.

New answer posted

7 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

x=qEt²/2m, y=gt²/2. y= (mg/qE)x. Straight line.

New answer posted

7 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

E = P? A? T? [ML²T? ²] = [MLT? ¹]? [L²]? [T]?
a=1, 2b+a=2⇒b=1/2, -a+c=-2⇒c=-1. [PA¹/²T? ¹].

New answer posted

7 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Kindly considere the following Image 

 

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

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 687k 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.