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\(Q = 4 sin^2\pi t/6 + 6 cos\,\pi t /4\)

\(i = \frac{d\theta}{dt}\)

\(i = \frac d{dt} (4 sin^2\pi t/6+ 6cos\,\pi t/4)\)

\(i = 4 \times 2sin \frac{\pi t}6 \times cos \frac{\pi t}6 \times \frac \pi 6 + 6x - sin \frac{\pi t }4 \times \frac \pi 4\)

\(i = \frac{8\pi}6 sin \frac{\pi t}6 cos \frac{\pi t}6 - \frac{6\pi}4sin \frac{\pi t}4\)

\(i = \frac {8\pi}6 \times sin(\frac{\pi \times 4}6) cos (\frac {\pi\times 4} 6) - \frac {6\pi }4\times 0\)

\(i = \frac{8\pi}6 \times \frac {\sqrt3}2 \times \frac{-1}2\)

\(= + \frac{\pi\sqrt 3}3\)

\(i = + \frac\pi{\sqrt 3}\) ampere 

Correct option is: (c) All bulbs glow with equal brightness

Correct option is: (c) earth

Given: 

Current (I) = 0.4 A 

Time (t) = 5 min = 5 x 60 = 300 s 

To find: 

Charge (Q) =? 

Formula: Q = 1 x 

Solution: 

Q = 0.4 x 300 

Q= 120 C. 

Charge passing through the conductor is 120

Correct option is: (b) potential

In the absence of any external source the motion of electrons in a conductor is random and electrons collide continuously with the positive ions of metal. This causes a random change in direction of motion. The average velocity of random motion of electrons in any direction is zero, hence current is zero.

When the bridge is balanced, there will be no current in key, therefore a constant current will flow in the galvanometer. In balanced position, there will be a constant deflection in galvanometer and hence no change in its deflection on pressing the key.

1. The value of unknown resistance X, may not be accurate due to nonuniformity of the bridge wire and development of contact resistance at the ends of the wire. 

2. To minimise these errors, the value of R is so adjusted that the null point is obtained in the middle one-third of the wire (between 34 cm and 66 cm) so that the percentage errors in the measurement of I_X and I_R are minimum and nearly the same.

The values of the two resistances R₁ and R₂ are 40Ω and 40Ω.

(a) Terminal voltage is less than the emf : Terminal Voltage < e.m.f.

(b) e.m.f. is equal to the terminal voltage when no current is drawn.

The internal resistance of a cell depends on :

1. Nature of the electrolyte : The greater the conductivity of the electrolyte, the lower is the internal resistance of the cell.

2. Separation between the electrodes : The larger the seperation between the electrodes of the cell, the higher is the internal resistance of the cell. This is because the ions have to cover a greater distance before reaching an electrode.

3. Nature of the electrodes.

4. The internal resistance is inversly proportional to the common area of the electrodes dipping in the electrolyte.

It can be seen that resistances BC and CD are in series and their combination is in parallel with AD.

Then \(\frac{1}{R_p}=\frac{1}{6}+\frac{1}{3}\)

\(\Rightarrow\) R_p = \(2\Omega\)

Total resistance of circuit is 2 + 3 = 5Ω

(Due to capacitor, resistor 3Ω in EF will not be counted)

Total current = \(\frac{15}{5}=3A.\)

This current gets divided at junction A.

Voltage across DF = 3 Ω × 3 A = 9 V and Voltage across AD = 15 – 9 = 6 V

I across CD = \(\frac{6\,V}{3\,+\,3}=1\,A\)

Hence, current through arm CD = 1 A.

Resistance of a conductor of length l, and radius r is given by

\(R=p\frac{l}{\pi r^2};\)

thus \(R=\infty\,\frac{1}{r^2}\)

V = ∈ - Ir \(\Rightarrow\) r = \(\frac{\epsilon\,-\,V}{I}\)

At I = 0, V = ∈

When V = 0, I = I0, r = \(\frac{\epsilon}{I_0}\)

The intercept on y-axis gives the emf of the cell. The slope of graph gives the internal resistance.

Correct answer is (b) π

The resistance of 1m wire is \(2\Omega\)

The circumference of the circle is \(2\pi r\)

Radius r= 1m

So, The total length is \(2\pi\)

Resistance of the wire is \(4\pi\Omega\)

So, Resistance of each segment is \(2\pi\Omega\)

R_eff = \(\frac{2\pi \times 2\pi}{2\pi + 2\pi}\)= \(\frac{4\pi^2}{4\pi}=\pi\)

No, the resistance remains same, because length and cross-sectional area of the wire remain unchanged.

A_A : A_B = 1 : 6

H = V²t/R and \(R=\frac{pl}{A}\)

\(H_A=\frac{V^2t}{pl/A_A};\)  \(H_B=\frac{V^2t}{pl/A_B}\)

\(=\frac{H_A}{H_B}=\frac{V^2\times A_A}{V^2tA_B}\)

\(\Rightarrow \frac{H_A}{H_B}=\frac{A_A}{A_B}=1:6\)

60W bulb is more resistance.

For same p.d. the heat produced per second

\(H=\frac{V_2}{R}\,∞\,\frac{1}{R}.\)

As the one part of heater coil has resistance R₂ = R/₂ being the resistance of original coil; therefore the ratio of heat produced

\(\frac{H_2}{H_1}=\frac{R_1}{R_2}=\frac{R}{R/2}=2:1\)

Let the resistance of the bulb be R

R = V²/ P

R = 12² / 36 = 4Ω 

Now, new V₁ = 20 

R₁ = V₁² / P

R₁ = 400/ 36 = 11Ω

Required resistance = 11Ω - 4Ω = 7Ω

Thus, 7Ω is required to glow the bulb with full intensity. 

V = IR

⇒ V = I(R₁ + R₂)

⇒ 20 = 3(4 + R₂)

⇒ 4 + R₂ = 6.7

⇒ R₂ = 6.7 - 4

⇒ R₂ = 2.7 Ω

Given :

Voltage of Bulb, V' = 12V

Power of Bulb, P = 36W

Potential of the Cell, V = 20V

To Find :

Resistance Required to Glow the Bulb with Full Intensity.

Resistance of the Given Bulb :

⇒ R₁  = (V')²/P

⇒ R₁  = (12)²/36

⇒ R₁  = 144/36

⇒ R₁  = 4Ω

Current passes through the Bulb :

⇒ I = P/V'

⇒ I = 36/12

⇒ I = 3A

It is a Series Connection. So that, The Current passes through the Whole Circuit is 3A .

Equivalent Resistance R = (R₁ + R₂)

Now, Substituting the Values in Ohm's Law :

⇒ V = IR

⇒ V = I(R₁  + R₂)

⇒ 20 = 3(4 + R₂)

⇒ 3R₂ = 20 - 12

⇒ R₂ = 8/3

⇒ R₂ = 2.67Ω

The Resistance of 2.7Ω is Required to Glow it with Full Intensity.

Kirchhoff’s current law is consistent with the conservation of electric charge while the voltage law is consistent with the law of conservation of energy.

Correct option is: (a) series

Correct option is: (b) parallel

R₁ =1+2 = 3 ohm

R₂=1.5 ohm

R₁ and R₂ are connected in parallel

1/R = 1/R₁ + 1/R₂ = 1/3 + 1/1.5 = 1

R = 1 oh m

1.  Current I = \(\frac{total\,voltage}{total\,resistance}\)

I = \(\frac{5}{2+3}\) = 1A

2.  Voltage across 2Ω

V = 1 × 2 

V₂ = 2 V

Voltage across 3Ω

V = 1 × 3 = 3V

3.  5V

4. 3V

The potential difference V= – I r. 

The terminal voltage of the cell is less than its emf as the cell emf which is brought in the dielectric in the direction opposite to direction of external terminal voltage.

When circuit is open,the terminal potential difference is equal to emf of the cell . When current is drawn from the cell, some potential drop takes place due to internal resistance of the cell. 

Hence terminal potential difference is less than the emf of a cell and is given by 

 V = E - Ir

An electric cell is a general device used to produce a uniform flow of charges in a circuit. Its main function is to maintain a constant potential difference between its two terminals.

Electric cells are used in radio sets, wall clocks, wrist watches, torches, toys, mobile phones, cars, invertors, remote controls, ships, submarines, satellites, etc.

(a) greater
(b) lower
(c) nearly independent of
(d) 10²²

The random motion of free electron gets directed towards the point at a higher potential.

In the absence of an electric field, the motion of electrons in a metal is random. There is no net flow of charge across any section of the conductor. So no current flows in the metal. 

It can be seen that for all the observations, the ratio of V to I is constant to fairly good accuracy i.e. ohm’s law is obeyed accurately. Hence, the resistivity of alloy manganin is nearly independent of temperature.

Ohm's law is obeyed only when the physical conditions and the temperature of the conductor remain constant.

Smaller the resistivity of a substance, larger is its conductivity. The resistivity of silver is least so silver is the best conductor.

Germanium and Silicon. 

The resistance-change factor per degree Celsius of temperature change is called the temperature coefficient of resistance. A positive coefficient for a material means that its resistance increases with an increase in temperature.

Metals – These are solids which have very low resistivity or very high conductivity). 

Hence, σ ~ 102 – 108 S/m 

ρ ~ 10-2 – 10-8 Ωm

Semiconductors: A semiconductor remains partially full valence band and partially full conduction band at the room temperature The energy gap is narrower. These are solids which have resistivity or conductivity values between those of metals and insulators. Hence, 

σ ~ 105 – 10-6 S/m 

ρ ~ 10-5 – 106 Ωm

Alloys are metallic compounds composed of one metal and one or more metal or non-metal element.

Examples :

Steel, a combination of iron (metal) and carbon (non-metal)

Steel, for example, requires the right combination of carbon and iron (about 99% iron and 1% carbon, as it turns out) in order to produce a metal stronger, lighter and more workable metal than pure iron.

Correct answer is (a) metal

In insulators and semiconductors, as temperature increases, resistance Decreases.

The unit of specific resistance is Ohm meter.

A material with a negative temperature coefficient is called a Thermistor.

The reciprocal of electrical resistivity is called Electrical conductivity.

Volta first introduced the electrochemical battery

(i) → (b) 

(ii) → (c) 

(iii) → (d) 

(iv) → (a)

Unit of electrical energy is Joule.

(i) → (d) 

(ii) → (c) 

(iii) → (b) 

(iv) → (a)

Correct answer is (a) j = σE