# 20 - Current Electricity Questions Answers

Please explain krotov questions 3.26 of electricity

**Asked By: RITESH KUMAR**

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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

**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

${\int}_{r}^{\infty}\frac{\rho dx}{4\pi {x}^{2}}=\frac{\rho}{4\pi}\left[-\frac{1}{x}\right]\phantom{\rule{0ex}{0ex}}onputtinglimitsresis\mathrm{tan}cearound\phantom{\rule{0ex}{0ex}}onesphere=k\frac{1}{r}\phantom{\rule{0ex}{0ex}}similarforotherspheresonet{R}_{1}=2\frac{k}{r}\phantom{\rule{0ex}{0ex}}Insecondcaseradiusofonesphere\phantom{\rule{0ex}{0ex}}ishalfmeansr/2\phantom{\rule{0ex}{0ex}}Thus{R}_{2}=\frac{k}{r}+\frac{k}{r/2}=3\frac{k}{r}\phantom{\rule{0ex}{0ex}}nowforsamebattery{i}_{1}{R}_{1}={i}_{2}{R}_{2}\phantom{\rule{0ex}{0ex}}solveandwaitforvideo\phantom{\rule{0ex}{0ex}}$

**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

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 = l^{2}/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

**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

**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

Two voltameters one of * copper* and another of

*, are joined in parallel. Whe n a*

__silver__*flows through the voltameters, equal amount of metals are deposited.*

__total charge q__If the

*which flows through the silver voltameter is*

__electrochemical equivalents of copper and silver are z1 and z2 respectively the__

chargecharge

**Asked By: JADAWALA UDIT**

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**Joshi sir comment**

m = z_{1}q_{1} = z_{2}q_{2} (1)

and q = q_{1} + q_{2}

now solve