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Ohm's Law is applicable to all linear or ohmic circuits, regardless of whether they are d.c. or a.c. A linear or ohmic circuit is one in which the ratio of voltage to current is constant for variations in voltage. Ohm's Law does not apply to non-linear or non-ohmic circuits (e.g. tungsten, electronic devices such as diodes, and electrolytes).

(i)  1Ampere of current means 1 Coloumb Of charge flowing through a conductor in 1 Second.

Also 1 Ampere of current May be defined as the current flowing through two parallel conductors in same direction when they are separated by 1 metre and they experiences a force of 1 N/cm by virtue of each other.

(ii) One volt is defined as energy consumption of one joule per electric charge of one coulomb. 1V = 1J/C.

OR

One Volt is defined as energy consumption of one joule per electric charge of one

coulomb.

(iii) Ohm's LAW : It states that "Physical condition remaining same, the current flowing through a conductor is directly proportional to the potential difference across its two ends". i.e., V = IR where the constant of proportionality R is called the electrical resistance or resistance of the conductor

(iv) Charge is the property of matter which enables it to experience a force when placed in an electric or magnetic field.

The SI unit of charge is coulomb (C). 
1C is the charge carried by  6.025 * 10^-19 electrons 

The answer is (d) the same in all the cases

There are no changes in any of the circuits, hence current will be same in all the circuits

The answer is (c) Brightness of bulb B will be more than that of A

• Bulbs are connected in parallel so resistance of combination would be less than arithmetic sum of resistance of all the bulbs. 
• So there will be no negative effect on flow of current. 
• As a result, bulbs would glow according to their wattage.

(i) Other bulbs will glow with the same brightness. 

(ii) When the bulbs are in parallel, wattage will be added (4.5 W) and the ammeter reading would be

I= P/V= 4.5 W/4.5V =1.0 Ampere

(iii) Since ammeter reading is 1.0 ampere, resistance of the combinatio., is 4.5/1.0 A=4.5Ω

The resistors R₁ and R₂ are connected in the parallel combination because the potential difference across both the resistors is same.

The bulbs will provide more illumination when connected in parallel as each bulb will get the required voltage/current and their illuminations will be added, In case the bulbs are connected in series, they will not get the required voltage and may not glow.

The bulbs should be connected in parallel because each bulb will show its maximum power or total resistance decrease hence greater current may be, drawn from the cell.

Advantages of parallel arrangement :

(i) The voltage across each device is same and it can take current as per its resistance.

(ii) If one device is damaged, then others continue to work properly.

By Ohm’s Law V = IR

⇒ \(R =\frac{V}I\) = slope of V-I graph 

Therefore, the slope of the V-I graph represents the effective resistance. 

Now in series combination of the resistance all the resistances get summed up 

R_eq = R₁ + R₂ + R_{3+ ………….+ Rn.} 

Therefore, the series combination will have the maximum resistance than the other resistors.

Hence the line with maximum slope (line C) will give the series combination of the other two resistors.

As R₁>R₂
Therefore, figure (a) is correct. This is because R₁ is greater so the slope of V-I graphs (V/I) is greater, in figure (a) and is correctly represented as R₁ .

Solution : 

Electric resistance.

The fuse of the device of rating 11W; 220V will burn.

R = V²/W,

Resistance oi 11W device will be four times than that of the device of 44W.

The voltage across 11W: 352V

Voltage across 44W = 88V

352V across the device of 11W; 220V rating sufficient to burn the fuse of the device.

Correct Answer is: (d) √(3λa/ɛ₀)

The maximum length of the string which can fit into the cube is √3a, equal to its body diagonal. The total charge inside the cube is √3aλ, and hence the total flux through the cube is √(3aλ/ɛ₀.)

The resistance of an ammeter is very low. If an ammeter is connected in parallel, the resultant resistance of the circuit decreases and excessive current passes through the instrument. Hence, the ammeter is likely to be burnt out

Correct answer is (d) cow-dung

Solution : 

(a) A long piece of nichrome wire. 

(b) A thin piece of nichrome wire.

Solution :

On decreasing the temperature, the resistance decreases. 

Solution :

Presence of impurities in a metal increases the resistance

Electric potential is the work done in carrying a unit positive charge from infinity to a point. If W is the work done q is the charge, then electric potential V = W/q

The SI unit of electric potential is Volts (V)

The electric potential difference between two points is the work done in carrying a unit positive charge from one point to the other.

The electric potential difference between points A and B V_AB = Work done to carry charge q from A to B / charge q

The SI unit of electric potential difference is Volts (V)

Electric potential energy is the work which has to be done to bring charges to their respective locations against the electric field with the help of a source of energy.

This work done is stored in the form of potential energy of the charges.

‘A’ shows parallel combination because in case of parallel combination, the resistance is less for their potential is less, currents is more. 

Earth wire – Insulation is of Green or Yellow colour.

An ohmic resistor is a resistor which obeys ohm’s law i.e., its resistance remains constant with increase in temperature 

V ∝ I 

V = IR at constant temperature.

The answer is (b) IR²

IR² does not represent electrical power in a circuit

Resistance is the opposition to the flow of current.

The SI unit of resistance is OHM (Ω)

1 ohm is the resistance offered by a wire carrying 1 A current when 1 V is applied across its ends.

In the flow of current in a conductor, electrons are drifted. During this drift or flow, electrons mutually collide against each other and the atoms of the conductor. These collison give rise to some obstruction to the flow of electrons. This opposition or obstruction offered by a conductor to the flow of electrons is called ‘Resistance’. According to Ohm’s law, the ratio between potential difference V and the current i, is a constant provided the temperature and physical conditions of the conductor remain unaltered. As such the ratio V/I is called the resistance, i.e., ‘Ohmic resistance’ of the conductor.

Fuse is a safety device which is used to limit the current in an electric circuit.

The reason is that when the cell is being used, there is a drop in potential across the internal resistance of the cell. 

There are two kinds of charges- positive and negative.

Like charges repel and unlike charges attract each other.

There are two main limitations: 

(i) Temperature, 

(ii) Physical condition of the conductor. It is found that the ratio V/I is mo longer constant when the temperature is not kept constant. Generally, resistance increases with the rise in the temperature of the conductor. 

Physical conditions of any conductor mainly include: 

(i) Its length, 

(ii) Its cross-sectional area, 

(iii) The kind of the material. If there is no change in any of the above three conditions and also other condition like temperature remains constant, then Ohm’s law holds good, i.e., the ratio V/I remains constant.

Potential at a point is defined as the work done in bringing a unit positive charge from infinity to that point. Its S.I. unit is volt, where 1V = 1joule/1coulomb

The resistance offered by a conductor of unit length and unit cross section is known as resistivity.

We now that,

P = \(\frac{V^2}{R}\)

R = \(\frac{V^2}{P}\)

Now, as all the bulbs are given the same voltage so resistance of the 40W bulb is the greatest.

As the bulbs are connected in series so the current through the bulbs are the same

P = i²R

So the power if the 40W bulb is the greatest and the power dissipated gives the glow of the bulb.

P=10W, V=220V, I=5A 

We know that 

P=VI =220X5 P=1100W 

Power of one bulb=10W 

Total no. of bulbs that can be connected=1100/10=110

(a) Lamp; because least current is flowing through it. 

(b) Large current drawn by the kettle; Earth connection needed. 

(c) We know that 

P=VI V=240 V, I=8.5 A P=240X8.5=2040 W=2.04 kW

(d) When connected to 240 V supply, P=2040W R = V₂/P = 2402/240 R=28.23ohm 

Now, when V=120 V, R=28.23ohm I=V/R=120/28.23=4.25 A