Electricity

The SI unit for measuring an electric current is the ampere, which is the flow of electric charge across a surface at the rate of one coulomb per second. Electric current is measured using a device called an ammeter.

To find the minimum resistance we have to join all the resistance in parallel

∴

⇒

⇒

⇒ Ohm

Minimum Resistance is 0.04 Ohm.

Figure below show the Schematic diagram for the same

We know that H = VIT

And V = IR by Ohm’s law

⇒ H = (IR)IT

⇒ H = I²RT

This is the Mathematical form of the Joule’s law of heating effect.

Iron and heater are the devices that use the Joule’s law of heating effect.

Resistance of an ammeter should be low. If the resistance is more the current will be less. If the ammeter has zero resistance then we will get the exact current but it is not possible as every material in the every material have some resistance. So the resistance of the ammeter should be low.

By Ohm’s Law

Case 1:

Resistance(R) =

Case 2:

Resistance(R) =

Case 3:

Resistance(R) =

Case 4:

Resistance(R) =

Case 5:

Resistance(R) =

According to the Ohm’s Law the ratio of the current to the potential difference is always constant (here it is 25). But as we can see in Case 3 the ratio is 3 Ohm while in all other cases it is 25 Ohm.

∴ The trend where Potential Difference is 4.5 V and the current is 0.15 A is different.

Mean Value Of resistance =

= 25 Ohm.

Mean value of resistance is 25 Ohm.

We Know that

Resistance (R) =

⇒ Resistance (R_A) =

⇒ Resistance (R_B) =

∵ 1,210 (R_A) > 806.67 (R_B)

∴ Electric Bulb A has the greater resistance

a. Conductor has very low resistance in order to allow the free flow of electrons through them. Therefore Material A is the best conductor because it has the low resistivity which implies low resistance.

Insulators are the materials having high resistance. Therefore Material B is the insulator because it has the highest resistivity which implies high resistance.

b. The Resistivity of the wire will not change with the change in length. Resistivity is a property of the wire, it doesn't depend upon it’s length or cross-sectional area.

If we connect the given three resistors as shown in the figure below then we get net resistance as 1Ω. Here all the resistors are connected in parallel.

∴their equivalent resistance will be given as

⇒

⇒ Req = 1 Ω.

Therefore, the resistances must be connected in parallel to get equivalent resistance of 1 Ω.

We have,

Length of the metal wire, l = 1 m

Diameter of the wire, d = 0.3 mm

Radius, r = 0.3/2 mm = 0.15 mm

= 0.15 x 10^-3 m

Resistance of the metal, R = 25 Ω

Thus,

Resistivity (ρ) =

⇒ ρ =

⇒ ρ =

⇒ ρ = 1.9625 × 10^-6 Ωm

The resistivity of the metal is 1.9625 × 10-6 Ωm

An ammeter is used to measure the current flowing through a component/circuit. Since, current remains same in series connection and also the resistance of an ammeter is very small due to which it doesn’t affect the current to be measured, hence ammeter is connected in series with the circuit. Figure below shows the connection of the Ammeter in the circuit.

A voltmeter is used to measure the potential difference across two point in a circuit. Since, in parallel connection the voltage in the branches remains same, and also the resistance of the voltmeter is very high due to which a very small current flow through the voltmeter, therefore voltmeter is connected in parallel to the circuit.. Figure below shows the connection of the Voltmeter in the circuit.

The heating element is made up of alloy which has very high resistance so when current flows through the heating element, it becomes too hot and glows red. But the resistance of the cord of the heater which is usually of copper or aluminium is very low so it does not glow.

i. The two resistors are in the parallel combination because different amount of current flows through each resistor and each resistor has the same potential difference.

ii. Since the potential difference remains the same in the parallel combination

∴ V = 3 V

R = 10 Ω

By Ohm’s law

Current through 10 Ω resistance is 0.3 A

iii. Current for 15 Ω resistor is

Reading of Ammeter = Sum of current through 10 Ω and 15 Ω resistor = 0.3 A + 0.2 A = 0.5 A

Ammeter reading is 0.5 A

Two safety measures commonly used in electric circuits and appliances are

1. Electric Fuse: An electric fuse is connected in series it protects the circuit from overloading and prevents it from short circuiting.

2. Proper earthing of all electric circuit in which any leakage of current in an electric appliance is transferred to the ground and people using the appliance do not get the shock.

Following precautions should be taken to avoid the overloading of domestic circuits:

1. Do not use too many appliances at the same time.

2. Use the appliances within the safe limit of electric circuit.

An electrical fuse works by breaking the circuit when there is a fault in an appliance that causes too much current to flow. The wire within the circuit melts due to the heat generated when the current going through the circuit is too great.

As the current in a wire increases so does the heat generated. A fuse is simply a thinner piece of wire in the circuit. If the current gets too high the thin wire will melt and stop the current. The most commonly fuses used in the house are 3A, 5A and 13A.

Figure below shows the fuse wire in normal state and after being fused.

Using parallel arrangement in domestic wiring has many advantages:

1. Each electrical appliance will work under constant voltage because potential difference is same in parallel combination.

2. When two or more appliances are used at the same time, each appliance will be able to draw the current as needed.

3. When distribution circuits are in parallel, then each circuit operates separately. So, if one of the distribution circuits gets overloaded, only the fuse in that circuit will be blown off. And other distribution circuits will remain unaffected.

Ohm’s Law

Ohm’s law states that the current flowing through a conductor is directly proportional to the potential difference across the conductor. The proportionality constant is the resistance of the conductor; it is a constant for a known temperature range.

Verifying Experimentally:

1. Connect the wires as shown in the diagram.

2. Increase the rheostat to increase the current in the circuit.

3. Obtain the value of current in the circuit by the ammeter and also find the potential difference using the voltmeter.

4. For increase in current see the brightness of the bulb connected. Use your knowledge so far to obtain the resistance of the bulb.

Graph of Ohm’s law is a straight line obtained when current in the circuit (I) is plotted along x axis and potential (V) along y axis. The Graph obtained is shown in the figure below.

Conditions For Ohm’s Law to hold.

Ohm's law does not hold under all conditions.

1. Ohm's law does not hold for non-ohmic material such as electrolyte.

2. In varying temperature conditions ohm's law does not hold good.

Electrical Resistivity: Electrical resistivity means opposition of flow of electric charges. Its S.I. Unit is OHM denoted by Ω named after the scientist Georg Simon Ohm in 1827.

Experiment:

Aim: To study the factors on which resistance of conducting wires depends.

Apparatus Required: A cell, an ammeter, nichrome wires of different length but same area of cross-section (thickness), nichrome wires of same length but different thickness, copper and iron wire of the same length and same thickness as that of any nichrome wire.

Procedure:

1. Connect the cell, an ammeter and plug key in series with nichrome wire in the gap XY as shown in the figure below.

2. Close the key and note the reading of ammeter. It measures the current ‘I1’ through the nichrome wire (marked ‘1’).

3. Replace the marked 1 wire with another nichrome wire having same area of cross-section (thickness) but of double length’2l’ (marked 2).

4. Note the ammeter reading (I2) again after closing the key.

5. Again replace the marked 2 wire with marked 3 wires which have the same length but are thicker than marked 1 and 2 nichrome wires. Again note down the current (I3) through this wire.

6. Unplug the key. Remove marked 3 nichrome wire from the gap XY. Connect the copper wire marked 4 having same length and same area of cross-section as that of nichrome wire marked 1. .

7. Plug the key again and note the ammeter reading. It measures the current (I4) through copper wire.

8. Repeat the experiment with iron wire and measure the current (I5).

Observation:

1. Current ‘ I’ is half ofI1 i.e.,

2. Current I3 increases when thicker wire of same length and of same material
i. e., nichrome is used.

3. Current I₄ and I₅ is different for copper and iron wire.

Conclusion:

1. Different wires drew different amount of current from the same cell.

2. First observation indicates that the resistance of the conductor increases with increase in length. So, resistance is directly proportional to length.

3. Second observation shows that thicker wires have lesser resistance. So, resistance is inversely proportional to area of cross section of the wire.

4. Third observation shows that resistance of the conductor depends on the nature of its material.

We will do the following activity to conclude that the same voltage exists across three resistors connected in parallel.

Connect the three resistors in parallel across the battery as shown in the figure below

● Now disconnect the voltmeter and connect it across R1.

● Note the reading of voltmeter. It is found to be V.

● Disconnect the voltmeter and connect it across R2.

● Again disconnect the voltmeter and connect it across R3. Note the reading of voltmeter. It is also found to be V.

Conclusion: When resistors are connected in parallel to each other, potential difference across each resistor is same.

i. All bulbs will not glow with the same brightness. In series, they will glow dull as all the three bulbs do not get the equal potential difference, but as they get it in parallel, they will glow brighter in parallel.

ii. If one of the bulb fuses, all will glow continuously in parallel as they all have a different circuit, and their circuit is still active, but in series all will stop glowing because the circuit will break after one bulb is fused.

Below is the procedure to verify experimentally that the same current flows through every part of the circuit containing three resistances in series.

Procedure:

1. Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key. As shown in the figure below.

2. Plug the key. Note the ammeter reading.

3. Change the position of ammeter to anywhere record the reading each time.

We will see that there is no change in the reading of the ammeter.

Hence, the same current flows through every part of the circuit containing three resistors in series connected to a battery.

Joule’s heating effect : The Joule's law states that the quantity of heat produced in a resistor is directly proportional to:

(i)The square of current for a given resistance,

(ii)The resistance for a given current,

(iii)The time for which the current flows through the resistor,

⇒ H = I²Rt.

A simple experiment to demonstrate Joule’s heating effect is that if we switch on the electric bulb for a long period of time than it becomes hot.

Four applications of Joule's heating effect in daily life are:

(i) Electric fuse is a safety circuit devices work on this principle.

(ii) Electric iron we use to iron our clothes works on this principle.

(iii) Electric kettle used to boil water also works on this principle.

(iv) Electric toaster to make bake breads works on the same principle.