20 - Current Electricity Questions Answers

Physics >> Electricity >> Current Electricity IIT JEE

Please explain krotov questions 3.26 of electricity

Asked By: RITESH KUMAR

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Asked By: RISHI

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Joshi sir comment

Physics >> Electricity >> Current Electricity IIT JEE

Sir, in the formula for current , in the ncert it first talks about steady state current. Then it writes i=q/t. but then it generalises and writes the formual i=(delta)q/(delta)t

sir my doubt is that how do we define flow of charge in steady state across a cross section. and why is there (delta)q in formula since the charge is not changing with time.

is it the amount of charge passing per unit time or the change in charge per unit time?

pls help

Asked By: SAM LUTHRA

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Joshi sir comment

It is the amount of charge passing through a particular cross section of wire in unit time

Physics >> Electricity >> Current Electricity IIT JEE

Asked By: KUNDAN RAI

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Physics >> Electricity >> Current Electricity IIT JEE

Sir how to solve? My main problem is how is current flowing between the spheres? Is all the current coming out from one sphere going to the other sphere? Sir can you show with a diagram where exactly all the current is going? Thank you sir.

Asked By: DARIUS

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Joshi sir comment

Current comes out from the surface of one sphere symmetrically

and enters to the surface of other symmetrically.

Resistance of the medium around one sphere is

$\int_{r}^{\infty} \frac{ρ d x}{4 π x^{2}} = \frac{ρ}{4 π} [- \frac{1}{x}] o n p u t t i n g l i m i t s r e s i s \tan c e a r o u n d o n e s p h e r e = k \frac{1}{r} s i m i l a r f o r o t h e r s p h e r e s o n e t R_{1} = 2 \frac{k}{r} I n s e c o n d c a s e r a d i u s o f o n e s p h e r e i s h a l f m e a n s r / 2 T h u s R_{2} = \frac{k}{r} + \frac{k}{r / 2} = 3 \frac{k}{r} n o w f o r s a m e b a t t e r y i_{1} R_{1} = i_{2} R_{2} s o l v e a n d w a i t f o r v i d e o$

Physics >> Electricity >> Current Electricity IIT JEE

Any method that can be used other than superposition principle?

Asked By: LUFFY

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Joshi sir comment

By superposition principle, it is solved in minimum time and effort, other method is very complex. So use superposition principle.

Divide current in point 1 and 2, considering symmetry throughout, we get the equivalent resistance of the whole network except resistor R between 1 and 2 is 2R

Now 2R and R in parallel between 1 and 2

Now solve

Physics >> Electricity >> Current Electricity Medical Exam

if an area of isoscale right angle triangle(made of uniform wire of unit resistivity) is 8 unit. a battery of emf 10 unit is connected with across the longest branch then the current drawn from battery is approx ans is 3A

Asked By: AKANKSHA SINHA

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Joshi sir comment

area = l²/2 = 8 so l = 4

ρ = 1

so resistance of the 3 legs of the triangle = 4/A, 4/A, 4√2/A

battery is connected across the longest branch so eq. resistance = 4/A+4/A+4√2/A = 13.6/A

i think area of cross section will be given then calculate current

Physics >> Electricity >> Current Electricity IIT JEE

Can You explain some easy way to find potential between A and B?

Asked By: BONEY HAVELIWALA

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Joshi sir comment

use kirchhoff laws

4-10i+1-30x = 0

1-30x+20(i-x) = 0

solve it for i and x

potential across A and B = 4-10i+1 = 30x

Physics >> Electricity >> Current Electricity IIT JEE

Find The equivalente resistance between A and B. Each resistor is of 4 ohm.

Asked By: BONEY HAVELIWALA

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Joshi sir comment

since the circuit is symmetrical with respect to the line given so V, the potentials at the given points will be same

now calculate the eq. for the half circuit and double it for getting the eq. of the whole circuit

Solution by Joshi sir

since the circuit is symmetrical with respect to the line given so V, the potentials at the given points will be same

now calculate the eq. for the half circuit and double it for getting the eq. of the whole circuit

for one side eq. of 2 and 4 = 4/3

and then 4 in series of it so 4/3 + 4 = 16/3

finally 16/3||4/3

eq = 16/15

final answer = 16/15 * 2 = 32/15

Physics >> Electricity >> Current Electricity AIEEE

Two voltameters one of copper and another of silver, are joined in parallel. Whe n a
total charge q flows through the voltameters, equal amount of metals are deposited.
If the electrochemical equivalents of copper and silver are z1 and z2 respectively the
charge which flows through the silver voltameter is

Asked By: JADAWALA UDIT

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Joshi sir comment

m = z₁q₁ = z₂q₂ (1)

and q = q₁ + q₂

now solve

20 - Current Electricity Questions Answers

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