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(a) volts; voltmeter; parallel

(b) conductor; insulator

Solution : 

(a) Potential difference = Work done/Charge moved. 

(b) V₁=220 V, V₂=230V, Charge moved=4C 

Thus, the potential difference= V₂- V₁ =230-220 =10. 

We know that, Work done = Potential difference x Charge moved = 10 x 4 Work done = 40 joules 

Solution :

Potential difference between two points in an electric circuit is defined as the amount of work done in moving a unit charge from one

One coulomb of charge is that quantity of charge which exerts a force of 9 × 109 Newton on an equal charge is placed at a distance of 1 m from it.

The electric potential (or potential) at a point in an electric field is defined as the work done in moving a unit positive charge from infinity to that point. Unit of electric potential is volt.

The unit of electric charge is Coulomb.

Voltmeter the potential difference across a conductor.

Solution : 

(a) Voltmeter 

(b) Given : Potential difference=12V, Charge moved=1C 

We know that, Work done = Potential difference x charge moved = 12 x 1 = 12 joules 

Since work done on each coulomb of charge is 12 joules, the energy given to each coulomb of charge is also 12 joules. 

The SI unit of potential is Volt.

(a) p.d. stands for potential difference.

(b) Voltmeter is used to measure p.d.

The Correct option is (C) 1.6 A 

Here,

I₂ = 120cm = 1.20m 

R₁ = 5.0Ω 

R₂ = 2.5Ω 

Now,

R₁ = \(\frac{ρ\,\times\,I_1}{A}\)

And,

R₂ = \(\frac{ρ\,\times\,I_2}{A}\)

(Here, resistivity ρ and the area A remain same as the wires are similar.)

∴ \(\frac{R_1}{R_2} = \frac{ρ\,\times\,\frac{I_1}{A}}{ρ\,\times\,\frac{I_2}{A}}\)

\(\frac{R_1}{R_2} = \frac{I_1}{I_2}=\frac{I_1}{120\,\times10^{-2}}\)

Now,

\( \frac{5.0}{2.5}=\frac{I_1}{120\,\times10^{-2}}\)

l₁ = 2×120×10-2 m 

l₁ = 24×10-1 m 

∴ l₁ = 240 cm

R= V/I, R = 2/0.1 = 20Ω

(i) (a) In (I) current is constant while in (II) current is variable.

(b) In (I) current is in the positive state while in (II) current is in both states.

(ii) l-Direct current

II - Alternating current

(iii) I-from battery

II - From power supply in our houses

(iv) Because current type is alternating current that's why frequency varies form 0 to 50 Hz.

The brightness of glow of bulb P will increase and brightness of glow of bulb Q will decrease. This is because on closing S, bulbs Q and R will be in parallel and the combination will be in series with bulb P. Hence the total resistance of the circuit will decrease and the current flowing in the circuit will increase. Therefore, the glow of bulb P will increase. Also since bulbs Q and R will be in parallel in this case, the current gets divided and lesser current flows through Q and hence the glow of bulb Q will decrease.

Solution : 

In case of parallel combination, the resultant resistance will be less than either of the individual resistances. 

F α (q1q2)/r²

The electrostatic force of attraction or repulsion between a pair of charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

F = (Kq1q2)/r²

K is the constant of proportionality and is equal to 9 × 10⁹ Nm²/C² for free space.

For similar charges the force is repulsive and for dissimilar charges it is attractive.

According to the Law of conservation of charge “When two different bodies are rubbed together, both bodies get charged equally but with charges of opposite kind.” Thus, the total charge of two bodies before and after rubbing remains the same. 

Example: When an ebonite rod is rubbed with fur, the electrons from the fur are transferred to the ebonite rod and therefore the ebonite rod becomes negatively charged (due to gain of electrons), while the fur becomes equally positively charged (due to a deficit of the same number of electrons). As the same no. of electrons are, exchanged by the ebonite rod and fur, the magnitude of charges main same but with opposite sign.

When an ebonite rod is rubbed with fur, the ebonite rod acquires negative charge and the fur acquires positive charge. This means electrons have moved from fur to ebonite. The net charge in the system remains the same. So charges are neither created nor destroyed but transferred from one material to the other.

Insulators are bad conductors of charges but they can be charged easily by friction.

Conductors allow free flow of charges.

Current is the rate of flow of charge.

If q is the charge in coulomb and t is the time is seconds then current I=q/t

The SI unit of current is ampere (A).

Current is a scalar quantity.

While connecting copper wire in the circuit, the copper coating at the two ends erf the copper wire should be cleaned or scratched.

(a) 

• Resistance of the bulbs in series will be three times the resistance of single bulb.
• Hence, the current in the series combination will be one-third compared to current in each bulb in parallel combination. 
• The parallel combination bulbs will glow more brightly. 

(b) 

• The bulbs in series combination will stop glowing as the circuit is broken and current is zero. 
• However the bulbs in parallel combination shall continue to glow with the same brightness.

It is a device which converts solar energy into electrical energy

It is a device which convert chemical energy into electrical energy.

The current flows in closed circuit from + ve to – ve terminal of cell

An electric cell has Two terminals.

Kilowatt hour (kWh) is the commercial unit for electrical energy

1 kWh = 3.6 × 10⁶ J

No. of units of electricity consumed in a household = no. of kWh

Total cost of electricity = total units × cost per unit of electricity

The wire which melts, breaks the circuit and prevents the damage of various appliances in household connections is called a fuse wire

A fuse wire is made of an alloy of aluminium, copper, iron and lead

The thickness of the fuse wire increases the maximum safe current that can flow through it

According to joule’s law of heating,

H = i² Rt

So, greater the resistance (i.e. greater resistivity), greater the heat produced.

To bear the amount of heat produced, the melting point of the element must be high.

Heat generated is given by,

H = \(\frac{V^2}{R}\,t\)

H ∝ V² 

Hence, with doubling of the potential difference, the heat produced will be four times.