# Short notes of SSC JE | Electrical Work, Power, And Energy

Q. 1. There is a frill of 40 bulbs (connected in series) in a room. One bulb is fused The remaining bulbs are again joined in series and ‘connected to the same supply. Will the light decrease or increase in the room?

Ans. Let R be the resistance of the whole ( 40 bulbs) frill. Power available from the whole frill is
P= V2/R
When one bulb is fused, the resistance of the frill is decreased and the power available from the frill is increased. Therefore, the light in the room increases.

Q. 2. The element of the heater is very hot while the wires carrying the current are cold. Why?

Ans. The element of the heater is made of nichrome while the wires carrying current are made of aluminum/copper. The resistance of nichrome is very high compared to that of copper/aluminum. Since it is a series circuit, the heat produced (H = I2Rt) depends only upon the resistance of the conductor in this case. Therefore, the element of the heater is very hot due to its high resistance while the lead wires are cold due to their low resistance.

Q. 3. What do you mean by the safe value of fuse wire current?

Ans. It is the maximum value of current that a fuse wire can carry without getting melted. Under such conditions, the rate of production of heat in the fuse wire is equal to the rate of loss of heat.

Q. 4. What is the difference between a heating wire and a fuse wire?

Ans. A heating wire is used in heating appliances whereas a fuse wire is used in a circuit to protect it from an overload. Therefore, the melting point of the material of the heating wire is high while the melting point of a fuse wire is low.

Q. 5. What are the factors upon which the safe current of a fuse wire depends?

Ans. The safe current of a fuse wire is given by; Ire r312/p. Therefore, the safe current of a fuse wire depends upon (i) the radius of the wire, and (ii) the resistivity of the material of the wire. It also depends upon the surrounding conditions such as temperature, etc.

Q. 6. Two bulbs of resistances 75 Q and 250 SI are connected in a room. Which will glow brighter? If one of the bulbs is turned off, what will be the effect on illumination?

Ans. All domestic appliances are connected in parallel so that the voltage across each bulb is the same. The rate of dissipation of energy, P = I/2/R. Therefore, the bulb of smaller resistance will glow brighter. When one of the bulbs is switched off, there will be no effect on the illumination of the other because, in a parallel circuit, the operation of each appliance is independent of the other. Therefore, the total light in the room will decrease.

Q. 7. Two bulbs, one of 60 W and the other of 100 W having the same voltage rating are connected in series across the supply in a room. Which bulb will glow brighter? If one of the bulbs is switched off, what will be the effect on illumination?

Ans. The 60 W bulb has more resistance (R = V2/P) than that of a 100 W bulb. Therefore, when the two bulbs are connected in series, the rate of dissipation of energy P(= I2R; I is the same) is more in that bulb which has larger resistance. Consequently, the 60 W bulb will glow brighter. When one of the bulbs is switched off, the circuit resistance decreases, and hence current increases. As a result, the rate of dissipation of energy P (= I2R) increases. Hence, the total light in the room will *increase.

Q.8. The nichrome heater consumes 1.5 kW and heats up to a temperature of 750°C. A tungsten bulb operating at the same voltage operates at a much higher temperature of 1600°C in order to emit light. Does it mean that the tungsten bulb consumes more power?

Ans. Not necessarily. The steady temperature attained by a resistor depends not only on the power consumed but also on its characteristics (e.g., surface area, emissivity, etc.) which decides heat loss due to radiation.

Q.9. A heater in series with a 60 W bulb is connected to the supply mains. If a 60 W bulb is replaced by a 100 W bulb, will the heater now give more heat, less heat, or the same heat?

Ans. In series combination, the rate of loss of heat is P (= 12R). When a 60 W bulb is replaced by a 100 W bulb, the circuit resistance decreases (•resistance of a 100 W bulb is 1 than that of a 60 W bulb). This increases circuit current and hence the rate of loss of I in the heater. Consequently, the heater will give more heat.

Q. 10. What are the important characteristics of a fuse?

Ans. A fuse has four important characteristics viz (I) physical size (ii) current rating (iii) voltage rating and (iv) fusing characteristic The current rating of a fuse should be more than the maximum normal circuit current by 25%. The voltage rating of the fuse must be at least equal to the circuit voltage. The fusing characteristic is determined by blowing time versus the current flow graph.

Q.11. how does a fuse protect electrical installation?

Ans. A fuse is connected in series with the circuit to be protected and carries the total circuit current. When a fault occurs (e.g. short circuit), the current through the fuse exceeds the permitted value. This raises the temperature and the fuse element melts or blows out. disconnecting the circuit protected by it.

Q. 12. An electric bulb rated for 500 IV and 100 17 is used in a circuit having a 200 V, supply Calculate the resistance R that one must put in series with the bulb so that the bulb delivers 500W.

Ans. The bulb will deliver 500 W if the voltage across it is 100 V and the current through it is I = 500/100 = 5 A. When the bulb is connected across a 200 V supply, let R be the required series resistance to meet these requirements. Now the circuit current is to be 5 A and the voltage across the bulb is to be 100 V. This means an extra 100 V (200 -100 = 100 V )must be dropped across R.

Value of R = 100/5 = 20 ohms