31 Years NEET Previous Year Questions: Current Electricity - 3 - Free MCQ

In series, R_s= nR

Since, the voltage is same for the two
combinations, therefore

H ∝ 1/R. 

Hence, the
combination of 39 bulbs will glow more.

For maximum current, the two batteries
should be connected in series. The current
will be maximum when external resistance is
equal to the total internal resistance of cells
i.e. 2Ω. Hence power devloped across the
resistacne R will be

To carry a current of 4 ampere, we need four
paths, each carrying a current of one ampere.

Let r be the resistance of each path. These
are connected in parallel. Hence, their
equivalent resistance will be r/4. According
to the given problem r/4 = 5 or r = 20Ω
For this propose two resistances should be
connected. There are four such
combinations. Hence, the total number of
resistance = 4 × 2 = 8.

At A current is distributed and at B currents
are collected. Between A and B, the
distribution is symmetrical. It has been
shown in the figure. It appears that current
in AO and OB remains same. At O, current i4
returns back without any change. If we
detach O from AB there will not be any
change in distribution.
Now, CO & OD will be in series hence its
total resistance = 2Ω
It is in parallel with CD, so, equivalent
resistance = 

This equivalent resistance is in series with
AC & DB, so, total resistance

Now 8/3 Ω is parallel to AB, that is, 2Ω so
total resistance

Between C & D, the equivalent resistance is
given by

Equivalent resistance along

∴ Effective resistance between A and B is

Faraday’s laws are based on the conversion
of electrical energy into mechanical energy;
which is in accordance with the law of
conservation of energy

In the electrolysis of CuSO₄, oxygen is
liberated at anode and copper is deposited
at cathode.

The two resistances are connected in series
and the resultant is connected in parallel with
the third resistance.

Voltage across 3 Ω resistance = 3 × 0.8 = 2.4V
This voltage is the same across 6 Ω
resistance. Hence current through this
resistance

Total current in the circuit
= 0.8 + 0.4 = 1.2 amp
Voltage across 4 Ω resistance
= 4 × 1.2 = 4.8 volts

The energy stored in the capacitor

This energy will be converted into heat in
the resistor.

The potential difference across 4Ω resistance
is given by

V = 4 × i₁ = 4 × 1.2 = 4.8 volt

So, the potential across 8Ω resistance is also
4.8 volt.

Current in 2Ω resistance i = i₁ + i₂
∴ i = 1.2 + 0.6 = 1.8 amp
Potential difference across 2Ω resistance
V_BC = 1.8 × 2 = 3.6volts

Power of heating coil = 100 W and voltage
(V) = 220 volts. When the heating coil is cut
into two equal parts and these parts are
joined in parallel, then the resistance of the
coil is reduced to one-fourth of the previous
value. Therefore energy liberated per second
becomes 4 times i.e., 4 × 100 = 400 J.

Initial resistance (R₁) = R; Initial length is
1and final length ( 2) = 0.5 . Volume of a
wire = . A. Since the volume of the wire
remains the same after recasting, therefore

We also know that resistance of a wire (R)

[Alt : When wires are drawn from same
volume but with different area of
cross-section, then

Length of each wire = ; Area of thick wire
(A₁) = 3A; Area of thin wire (A₂) = A and resistance of thick wire (R1) = 10 Ω

Resistance   (if is constant)

The equivalent resistance of these two
resistors in series
= R₁ + R, = 30 + 10 = 40Ω.

Correct Answer :- a

Explanation : We know that R_T = R₀(1+αT)  where RT = resistance at temp T, R₀ = resistance at temp T = 0^oC and α= temperature coefficient of resistance.

Here, 5 = R₀(1+α.50)...(1)

and 7 = R₀(1+α.100)...(2)

(2)/(1)

⇒ 7/5 = (1+100α)/(1+50α)

7+350α = 5+500α

150α=2

⇒α=2/150

=0.013/^oC

Equivalent Circuit

Equivalent Resistance of circuit

Given : Length of wire (l) = 15m
Area (A) = 6 × 10^–7 m²
Resistance (R) = 5Ω.
We know that resistance of the wire material

We know from the Kirchhoff 's first law that
the algebraic sum of the current meeting at
any junction in the circuit is zero (i.e. ∑i = 0)
or the total charge remains constant.
Therefore, Kirchhoff's first law at a junction
deals with the conservation of charge

Power = 100 W, Voltage = 200 V

Resistance of bulb = 

When bulb is applied across 160 V,
Current in bulb = 160/400 A

Power consumption = 

Resistance of ACB, R' = 3Ω + 3Ω = 6Ω.
For net resistance between A and B; R' = 6Ω
and 3Ω are in parallel.

In series, Equivalent resistance = 3R

∴ Equivalent resistance R' = R/3

m = Zit, 9 = Z × 10⁵, Z= 9 × 10–5
Again,
m = Zit = 9 × 10^–5 × 50 × 20 × 60 = 5.4 gm

Since H ∝ I² , doubling the current will
produce 4 times heat. Hence, the rise in
temperature will also be 4 times i.e., rise in
temperature = 4 × 5 = 20ºC.

In discharge tube the current is due to flow
of positive ions and electrons. Moreover,
secondary emission of electrons is also
possible. So V-I curve is non-linear; hence
resistance is non-ohmic.

In circuit, R_B and R_C are in series, so, Rs = 6 +
6 = 12Ω. This 12 Ω resistance is in parallel
with RA = 3 Ω,
So, equivalent resistance of circuit