In the electric equipment producing heat e.g., iron, electric heater, boiler, toaster etc., an alloy such as Nichrome is used, not pure metals. Give Scientific reason.

Resistivity of alloys like Nichrome is more than pure metals, so they will get heated easily when current flows through them. They have a high melting point and are stable at high temperature where they do not oxidise easily. So in the electric equipment used for heating e.g., electric iron, electric heater, boiler, toaster etc., an alloy such as Nichrome is used, and not pure metals.

Two tungsten bulbs of power 50 W and 60 W power work on 220 V potential difference. If they are connected in parallel, how much current will flow in the main conductor?

Given that wattages of two bulbs are P 1 = 50 W , P 2 = 60 W , potential difference of 220 V . We know that P = V I ----------- (1) Where P = power, V = potential difference, I = current Total power, P = 50 W + 60 W = 110 W Since bulbs are connected in parallel, the voltage drop will be equal. Now to find the total current I = ? From equation (1), I = P V Substituting, I = 110 W 220 V I = 0 . 5 A Therefore, the total current flowing through the main conductor is 0 . 5 A .

Observe the following diagram and choose the correct alternative.

The intensity of magnetic field in A is less than in B

The intensity of magnetic field in A is larger than in B

The intensity of magnetic field in B is less than in A

The intensity of magnetic field in A and B is same

Explain the construction and working of an electric motor.

Electric motors involve rotating coils of wire which are driven by the magnetic force exerted by a magnetic field on an electric current. When a rectangular coil is placed in a magnetic field and current is passed through it, a motor operates. The coil is subjected to a force that causes it to rotate continually. A magnetic field is created around the rectangular coil when electric current is conducted through it. An electric motor, as shown in the figure, consists of a rectangular coil ABCD of insulated copper wire. The coil is placed between the two poles of a magnetic field such that the arm AB and CD are perpendicular to the direction of the magnetic field. The ends of the coil are connected to the two halves P and Q of a split ring. The inner sides of these halves are insulated and attached to an axle. The external conducting edges of P and Q touch two conducting stationary brushes X and Y , respectively. Current in the coil ABCD enters from the source battery through conducting brush X and flows back to the battery through brush Y . Notice that the current in arm AB of the coil flows from A to B . In arm CD it flows from C to D , that is, opposite to the direction of current through arm AB . On applying Fleming’s left hand rule for the direction of the force on a current-carrying conductor in a magnetic field, we find that the force acting on arm AB pushes it downwards while the force acting on arm CD pushes it upwards. Thus the coil and the axle O , mounted free to turn about an axis, rotate anti-clockwise. At half rotation, Q makes contact with the brush X and P with brush Y . Therefore the current in the coil gets reversed and flows along the path DCBA . A device that reverses the direction of flow of current through a circuit is called a commutator. In electric motors, the split ring acts as a commutator. The reversal of current also reverses the direction of force acting on the two arms AB and CD . Thus the arm AB of the coil that was earlier pushed down is now pushed up and the arm CD previously pushed up is now pushed down. Therefore the coil and the axle rotate half a turn more in the same direction. The reversing of the current is repeated at each half rotation, giving rise to a continuous rotation of the coil and to the axle.

If a bulb of 60 W is connected across a source of 220 V , find the current drawn by it.

A potential difference of one Volt is equal to one Joule of energy being used by one Coulomb of charge when it flows between two points in a circuit. Given that, The power of the bulb is P = 60 W . The potential difference applied to the bulb is V = 220 V . The current flowing through the bulb is I = P V = 60 220 = 0 . 27 A .