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The power of a bulb is given by

P = \(\frac{V^2}{R}\)

Hence, the resistance will be minimum if the power is maximum. So, 60W bulb will have largest resistance.

Heat generated in an element is given by

H = \(\frac{V^2}{R}\,t\)

For largest amount of heat generation, R must be smallest and R is smallest in parallel connection.

As the equivalent resistance in series connection is the sum of the individual resistances, so it is greater than the largest resistance.

The potentials of the two conductors determine the direction of flow of electrons. The electrons flow from the conductor at low potential to the conductor at high potential.

Correct option is B) MCB

Correct option is C) Faraday
 

Correct option is C) 1000 W

A) Compact Fluorescent Lamp

Correct option is D) Nichrome

Correct option is D) fuse

Correct option is D) CFL bulb

Given: I = 0.5A, V = 3.0V

Resistance of filament of the bulb R = V/I = 3.0/0.5 = 6Ω

 Given: I = 100mA = 100 × 10^-3 A = 0.1A, e = 1.6 × 10^-19C

Let n electrons are passing per second. charge on one electron = e

Total charge passed in one second = ne

Thus, i = ne or n = i/e = 0.1/(1.6 x 10^-19) = 6.25 x 10¹⁷.

The formula of resistance is: R = \(\frac{ρ\,\times\,I}{A}\)

Where, ρ = resistivity 

I = Length of the conductor 

A = Area of the conductor 

∴ ρ = \(\frac{R\,\times\,A}{I}\)

Here: R = 10 Ω 

r = 0.25mm l = 75m  We know that the area of the wire will be: A = πr²

∴ A = 3.14 x (0.25 x 10^-3)²

A = 0.196 x 10^-6 m²

Putting the given values in the above formula

ρ = \(\frac{10\,\times\,0.196\,\times\,10^{-6}}{75}\)

ρ = 26.133 x 10^-6 Am

The main condition for the flow of charge between two points is the difference in their potentials. If the potential difference is greater, a strong current flows and if both points are at the same potential, no current will flow.

Metal body of an electrical appliance is earthed.

1. A cell 

2. Electrolyte 

3. Cathode 

4. positive 

5. chemical, electrical 

6. electrodes 

7. cells 

8 closed 

9. heat, light 

10. Compact Fluorescent Lamp 

11. Light Emitting Diode 

12. less 

13. Fuse 

14. Low 

15. Miniature Circuit Breaker

1. Wet stick contains some amount of water throughout it. 

2. Water is a good conductor of electricity (ofcourse, pure water does not pass current). 

3. If we touch the person to push away from the electric wire, the current passes through the water into our body. 

4. It causes to get electric shock to us also. 

5. So, we should not use wet stick to push away a person when he get electric shock. 

6. We should use dry stick only to do that.

(a) Electric power: It is the rate of doing work by an energy source or the rate at which the electrical energy is dissipated or consumed per unit time in the electric circuit is called electric power.

So, Power P = Work done (w)/Time (t)

= Electrical energy dissipated/Time (t)

= Vl = V²/R

(b) It means: the maximum current will flow through it is only 2 A. Fuse wire will melt if the current exceeds 2 A value through it.

(c) Given: p = 1 k W = 1000 W, V = 220 V

Current drawn, I = p/V = 1000/220 = 50/11 = 4.54 A

To run electric iron of 1 kW, rated fuse of 5 A should be used.

Electric current is a scalar quantity because it is the measure of how much charge flow through a particular area.

Smaller unit of current is Ampere denoted by A.

Charges can move if there is a difference of electric pressure or potential difference along the conductor.

Electric cell or a battery consisting of two or more cells.

The electric potential, or voltage, is the difference in potential energy per unit charge between two locations in an electric field.

I = nE/(nR + r) cells in parallel.

Given; 

Charge (Q) = 5C; 

Potential difference = 30 – 20 = 10V; 

Work done = Charge × Potential difference 

⇒ Work done = 5 × 10 = 50 J. 

Hence the work done is 50 Joule.

(i) P=1200 W

t= 30 min = 0.5 h

Electrial energy(E)= P x t

= 1200 x 0.5

= 600 W/h = 0.6 kWh

(ii) Energy consumed in 30 days

= 0.6 x 30 =18 kWh

Cost of using 18 unit = Rs. 4 x 18= Rs. 72

1/R = 1/r₁ + 1/r₂ + 1/r₃ resistances in parallel

R = r₁ + r₂ + r₃ , resistances in series.

The conductor of electricity is human body.

electrons are transferred from the glass rod to the silk

Resistance is inversely proportional to its cross section. As area is increasing resistance is less. Thus resistance of wire is changing with area of cross section.

(i) The conductivity of semiconductors increases when light falls on them and certain impurities are added to them.
(ii) Materials like Silicon, Selenium, Cadmium sulphide and Copper sulphide acts as semiconductors.

(i) If coils are connected separately V = 220 V

Resistance R₁ = 240

As per ohm’s law, V = IR

I = \(\frac{V}{R_1}=\frac{220}{24}\) = 9.166A

∴ If coils are connected separately 9.16A electricity flows in the coil.

(ii) If coils are connected in series

Resistance R₂ = 24Ω + 24Ω = 48Ω

As per ohm’s law V = IR

∴ I₂ = \(\frac{V}{R_2}= \frac{220}{48}\) = 4.58A

∴ If coils are connected in series 4.58A electricity flows.

Solution : 

Nichrome is an alloy of nickel, chromium, manganese ad iron having a resistivity of about 60 times more than that of copper. It is used for making the heating elements of electrical heating appliances.

P = \(\frac{V^2}{R}\)