15 - Current Electricity Questions Answers

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

 

  

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

 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

 

 

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 silver, are joined in parallel. When 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

Sir/Madam, 

                    Can you please explain me to find potential difference between two plates of capacitor like this shown in figure. Can you Give anymore examples if possible? Thank You! (Please omit time)

Asked By: BONEY HAVELIWALA
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Joshi sir comment

after time t charge on plates of capacitor q = CE(1-e-t/RC) 

then potential difference between the plates of capacitor = q/C

after a long time this potential difference = E      [this can be obtained by putting t = ∞]

Mam/Sir,  Please Help Me to Understand The Symmetry method And Star-Delta Method To find Eqivalent Resistance?

 

 

Asked By: GAUTAM SAGAR
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Joshi sir comment

fig 1 and fig 2 are explaining symmetry method. 

divide the circuit in two parts by a line symmetry, then calculate eq. resistance between A and O by series parellel rules then for getting eq. resisitance between A and B double the answer between A and O

 

Delta star is the method to convert the circuit into an equivalent circuit in which series parallel rules can be applied 

By this method a set of three resistances R1, R2 and R3 can be converted to 3 new resistances 1, 2 and 3

where 1 = R1R2/(R1+R2+R3), 2 = R1R3/(R1+R2+R3) and 3 = R2R3/(R1+R2+R3) 

In an electrical circuit...

H = W = i2Rt   [ H directly proportional to R ]

But i2Rt can also be written as (V/R) x (V/R) x R x t  =  V2t/R...

H = W =  V2t/R .  [ H inversely proportional to R ]

 

therefore my question is how can heat be both directly and inversely proportional to the resistance in the circuit??

Asked By: AAYUSH TANEJA
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Joshi sir comment

you should take the formula in which one variable is constant and for a circuit current will be constant

In an electrical circuit...

H = W = (i x i) x R x t   [ H directly proportional to R ]

H = W =  (V x V) x t /R .  [ H inversely proportional to R ]

So is Heat Generated directly proportional or inversely proportional to the resistance in the circuit ??

Asked By: AAYUSH TANEJA
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Joshi sir comment

H = W = (i x i) x R x t   [ H directly proportional to R ]

this one is correct for the heat generated

charge flowing thru a xsection of wire varies with time as q= 2te-kt , where k is const . the voltmeter reads zero at t=??

Asked By: SARIKA
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Joshi sir comment

q= 2te-kt 

on differentiating with respect to t, we get

dq/dt = 2[t(-k)e-kt e-kt ]

so for V = 0, i will be 0

so t = 1/k  

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