Electric Current - 7

Question :-
Electrical circuit is shown in figure. Calculate the potential difference across the resistor of 400 ohms as will be measured by the voltmeter V of resistance 400 either by applying Kirchoff's laws or otherwise.



Ans :- This circuit is equivalent to :



Which in turn is equivalent to :



As, it is a balance Wheat Stone's Bridge as
.
of 100 between A and B can be ignored.
The voltage will read the potential difference across Q.
First we find that i₁ = i₂ =
.
Potential difference across resistance Q = Q X i₁
.
Therefore, Voltmeter will read V.

Question :-
Charging and 100 V after charging. When charging below the current was 5A. What is the current at the end of the charging if the internal resistance is equal and constant to 8 in the whole process.

Ans :- The voltage supplied is constant for both times and equal V given by -

Let the current finaly (at the end of charging br I)


H -1: A conductors has a temperature independent resistance R and its total heat capacity is C. At t = 0 it is connect to d.c. voltage source V. Find T(t) assuing power dissipated into surrounding space to very as q = k(T - T₀) where k is constant, T₀ is surrounding temperature (and equals conductor's initial temperature) ?

Ans :- Energy supplied by source/unit time = energy lost in environment/unit time + energy used in raising temp/unit time


On solving : .

Dumb Question - 11. How we get the initial equation of the above question ?

Ans Initial equation states :
Energy supplied by source/unit time = energy lost to surroundings/unit time + energy used in raising temp/unit time
On close observation one finds that it is just the Law of Conservation of Energy.

M - 2. What amount of heat will be generated in a resistance R due to a charge q passing through it if the current in the coil :
(a) Decreases down to zero uniformly during a time internal to ?

Ans (a) Drawing corresponding i - t graph :




i₀ can be found by the information that charge q = area it

At any time t assume a small increament in time dt such that dt 0.
dH = Heat produced in this time interval = i²Rdt

Total Heat produced = .


Dumb Question - 18. How charge = Area under it graph ?

Ans :- We know that Current = I =
Thus, dQ = I dt
Integrating both sides
[Assuming at t = 0, Q = 0].
Now, is nothing but area under the It graph.

H - 3. A voltmeter of resistance R₁ and an ammeter of resistance R₂ are connected in series across a negligible resistance battery. When a resistance R is connected in parallel to voltmeter, ammeter increases its reading to three times and voltmeter's reading reduces by one third. Find R₁ and R₂ in terms of R ?

Ans :- Case I :


In this case E = I(R₁ + R₂).
Case II :



Now, as ammeter reads 3 times so, net current 3i is drawn from the battery.
Current in voltmeter resources to
Remaining passes through R.
V_C - V_D = R₁ = 8R.
Applying Kirchoff's Voltage Rule in ABFGA :



* Average Current
* Instantaneous Current
* Current Density
* Drift Velocity
* Ohm's law
* Resistivity
* Kirchoff's Laws
* Internal Resistance of a battery
* Grouping of Resistance :-
(a) Serial
(b) Parallel
* Heating Effects of Current
* Joule effect
* Maximum Power Transfer Theorem
* Grouping of Cells
* Galvanometer
* Ammeter
* Voltmeter
* Wheat Stone's Bridge
* Potentiometer