What is the meaning of the statement “The potential difference between two points is 1 V ” ? Name the device used to measure potential difference. What is resistance of a conductor? What is electric power? Write three formulae used to find electric power.

The potential difference between two points is said to be one volt ( 1 V ) , if 1 Joule ( 1 J ) of work is done to move a charge of 1 Coulomb ( 1 C ) from one point to another point in a current-carrying conductor. The device used to measure electric potential difference is voltmeter. The property of a conductor to oppose the flow of charge through it is called its resistance. The SI unit of resistance is Ohm ( Ω ) . The rate of electrical work done per unit time is called electric power. The following are the three formulae used to find electric power. 1) P = V I where ‘ V ’ is the potential difference in volts and ‘ I ’ is the current in amperes 2) P = I 2 R where I is the current and R is the resistance 3) P = V 2 R where V is the potential difference and R is the resistance

Observe the given circuit diagram. Calculate the total resistance and the total current flowing through the circuit.

Given, R 1 = 5 Ω , R 2 = 4 Ω , R 3 = 12 Ω , V = 24 V We have to find the total resistance of the circuit ( R T ) and the total current flowing through the circuit ( I ) . The total resistance of the circuit, R T = R 1 + [ 1 R 2 + 1 R 3 ] R T = 5 Ω + [ 1 4 Ω + 1 12 Ω ] R T = 5 + [ 3 + 1 12 ] R T = 5 + 4 12 R T = 5 + 1 3 R T = 5 + 3 [ 1 R p = 1 3 , R p = 3 Ω ] R T = 8 Ω According to Ohm's law, the total current flowing through the circuit, I = V R T I = 24 V 8 Ω = 3 A Therefore, the total current flowing through the circuit is 3 A .

How does overload and short-circuit occur in an electric circuit ? Explain. What is the function of fuse during this situation?

When too many electrical appliances of high power rating are switched on at the same time, they draw large current from the circuit. This is known as overloading the circuit. Overloading takes place when too many appliances are connected to single socket. This increases the flow of current and due to which the copper wires gets heated to very high temperature and catches fire. When the plastic insulation of the live wire and neutral wire gets torn, the two wires come in direct contact with each other. This is known as short-circuiting. During short-circuiting, when the two wires touch each other, a very small resistance between them is produced. Due to this, the current flowing through the wires becomes very large and heats up the wire resulting in fire. An electric fuse is a safety device which is connected in series in an electric circuit. It is a wire made up of a material having low melting point. When large amount of current flows through a circuit having a fuse wire, a large amount of heat is produced resulting in the melting of fuse wire. Thus, the circuit breaks up and the current stops flowing. This prevents the electric circuit from burning.

An electric refrigerator rated 400 W is used for 8 hours a day. An electric iron box rated 750 W is used for 2 hours a day. Calculate the cost of using these appliances for 30 days, if the cost of 1 kWh is Rs . 3 / - .

Given, Power of refrigerator, P r = 400 W The refrigerator is used for 8 hours a day. Total energy consumed by the refrigerator in 30 days = 400 × 8 × 30 = 96000 Wh = 96 kWh Cost of energy to operate the refrigerator for 30 days = 96 kWh × ₹ 3 per kWh = ₹ 288 Power of iron box, P i = 750 W The iron box is being used for 2 hours a day. Total energy consumed by the electric iron box in 30 days = 750 × 2 × 30 = 45000 Wh = 45 kWh Cost of energy to operate the electric iron box in 30 days = 45 kWh × ₹ 3 per kWh = ₹ 135 The total cost of using the refrigerator and iron box for 30 days = ₹ 288 + ₹ 135 = ₹ 423 Hence, the total cost of using these two appliances for 30 days is ₹ 423 .

According to Joule’s law of heating, mention the factors on which heat produced in a resistor depends. According to this law write the formula used to calculate the heat produced.

According to Joule's law of heating, the heat produced in a conductor is 1) directly proportional to the square of the electric current flowing through it, H ∝ I 2 2) directly proportional to the resistance of the conductor for a given current, H ∝ R 3) directly proportional to the time for which the electric current flows through the conductor, H ∝ t On combining the above three equations, we get H ∝ I 2 R t ⇒ H = k I 2 R t where k is the proportionality constant. If k = 1 , then H = I 2 R t Thus, the formula to calculate heat produced is H = I 2 R t (in Joules)

Draw the diagram of an electric circuit in which the resistors R 1 , R 2 and R 3 are connected in parallel including an ammeter and a voltmeter and mark the direction of the current.

The following is the diagram of an electric circuit with three resistors connected in parallel. In this diagram, A represents ammeter and V represents voltmeter. I represent the current flowing in the circuit. When the resistors R 1 , R 2 and R 3 are connected in parallel, the equivalent resistance is given by, 1 R p = 1 R 1 + 1 R 2 + 1 R 3 Here, the main current divides itself into different currents flows through each resistor. This means that, the currents in a resistor is inversely proportional to its resistance. The potential difference across each resistor is the same and is equal to that across the combination.