Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance?
Internal resistance of the circuit is negligible.
So, total resistance is R.
Therefore, current across the circuit is given by,
A capacitor, made of two parallel plates each of plate area A and separation d, is being charged by an external ac source. Show that the displacement current inside the capacitor is the same as the current charging the capacitor.
The charge on the plates is because of the conduction current flowing in the wires.
According to Maxwell’s equation, displacement current between the plated is given by,
where, is the lectric flux.
Now, using Gauss theorem,
Therefore, both the displacement current and conduction currents are equal.
Explain the term ‘drift velocity’ of electrons in a conductor. Hence obtain the expression for the current through a conductor in terms of ‘drift velocity’.
The velocity gained by the accelerating electrons in uniform electric field inside the conductor is drift velocity. The average velocity, acquired by free electrons along the length of a metallic conductor, due to existing electric field is called drift velocity.
Let ‘n’ be the number density of free electrons in a conductor of length ‘l’ and area of cross-section ‘A’.
Total charge in the conductor, Q = Ne
Time taken, t is given by,
Therefore, the current flowing across the conductor is given by,
, which is the amount of current flowing through a conductor in terms of drift velocity.
Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell.
Potentiometer can be used to measure the internal resistance of the cell.
A cell of emf E is connected across the resistance box through key K1.
When key K1 is opened galvanometer shows deflection at the balancing length l1.
So, E = k
If both keys are closed, then balancing point is obtained at length l2 (l2 < l1).
So, V = k
Now, using the relation,
Therefore, we have
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
Capacitance having a dielectric of thickness ‘t’ is given by,
When the thickness of the plates is reduced to half, t = d/2 then,
The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping?
The changing magnetic flux through the plate produces a strong eddy current in the direction, which opposes the cause as the plate oscillates.
Two charges of magnitudes – 2Q and + Q are located at points (a, 0) and (4a, 0) respectively. What is the electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin?
Using gauss law, we have
Electric flux is given by,
How does the mutual inductance of a pair of coils change when,
(i) distance between the coils is increased and
(ii) number of turns in the coils is increased?
Using the formula,
i) Mutual inductance decreases when the distance between the coils is increased.
ii) Now using the formula,
On increasing the number of turns in the coil, mutual inductance also increases.
Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic radiations. Name the radiations and write the range of their frequency.
The radiations are ultraviolet radiations.
The frequency range of UV rays is 1015 –1017Hz.
(a) For a given a.c., show that the average power dissipated in a resistor R over a complete cycle is
(b) A light bulb is rated at 100 W for a 220 V a.c. supply. Calculate the resistance of the bulb.
a) Average power consumed in resistor R is given by,
b) In case of ac, we have