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