It the resistence of an electic earn is 50 M and a current of 3.2 A how's through the restance.
Answers
Answer:
.
Q16. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Ans. Advantages of parallel connection are:
(a) In parallel circuit, if one electric appliance stop working due to some defect, then all other appliances keep working normally.
(b) In parallel circuit, each electrical appliance has its own switch due to which they can be turned on or off without affecting other appliances.
(c) Each electrical appliance get the same voltage (220 V) as that of the power supply line.
(d) In parallel connection of electrical appliances, the overall resistance of the circuit is reduced due to which the current from the power supply is high.
Q17. How can three resistors of resistances 2 Ω , 3 Ω , and 6 Ω be connected to give a total resistance of (a) 4 Ω , (b) 1 Ω?
Ans. R1 = 2 ohm
R2 = 3 ohm
R3 = 6 ohm
(a) When R2 and R3 are connected in parallel with R1 in series we get
1/R = 1/R2 + 1/R3
g= 1/3 + 1/6
= 1/2
Thus, R = 2 ohm
Resistance in series = R + R1
= 2 + 2
= 4 ohm
(b) When R1,R2, R3 are connected in parallel we get
1/R = 1/R1 + 1/R2 + 1/R3
= 1/2 + 1/3 + 1/6
= 1 ohm.
Q18. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω , 8 Ω , 12 Ω , 24 Ω?
Ans. (a) The highest resistance is when the resistances are connected in series:
R1 = 4 ohm R2 = 8 ohm
R3 = 12 ohm R4 = 24 ohm
Total resistance in series = R1 + R2 + R3 + R4
= 4 + 8 + 12 + 24
= 48 ohm
Thus, highest resistance is 48 ohm.
(b) The lowest resistance is when the resistances are connected in parallel
Total resistance in parallel
= 1/R1 + 1/R2 + 1/R3 + 1/R4
1/R = 1/2 + 1/8 + 1/12 + 1/24
= 12/24
1/R = 1/2 ohm, R = 2 ohm
Thus, lowest resistance is 2 ohm.
Q19. Why does the cord of an electric heater no, glow while the heating element does?
Ans. The resistance of cord is extremely small as compared to that of heating element, so the heat produced in cord is less as compared to heating element. So the heating element begins to glow but cord does not glow.
Q20. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Ans. Charge = 96000 coulomb
Time = 1 hour = 3600 seconds
Potential difference = 50 V
I = Q/T
I = 96000/3600 = 80/3 amp
V = I × R
50 = 80/3 × R
Thus. R = 15/8 ohm
Heat = I2RT
= (80/3)2 × 15/8 × 3600
= 800 × 6000
= 4800000 joules.
Q21. An electric iron a, resistance 20 Ω takes a current of 5 A Calculate the heat developed in 30 s.
Ans. Resistance of the iron = 20 ohm
Current = 5 amp
Time = 30 seconds
Heat produced = I2RT
= 52 × 20 × 30
= 15000 joules
Q22. What determines the rate at which energy is delivered by a current?
Ans. Electrical power determines the rate at which the energy is delivered by a current.
Q23. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2h.
Ans. Current in the motor, I = 5 amp
Potential difference, V = 220 V
Time = 2 hours
Power of the motor = V × I
= 220 × 5
1100 watt or 1.1. kWh
Energy = power × time
= 1.1 kWh × 2 h.
= 2.2 kWh
QUESTIONS FROM NCERT TEXTBOOK
Q1. A piece of wire of resistance R is cut into five equal parts. ‘These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then what is the ratio R/R’?
Ans. 25 : 1
Q2. Which of the following terms does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) W
(d) V2/R
Ans. (b) IR2
Q3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, power consumed will be—
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Ans. (d) 25 W.
Q4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be—
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Ans. (c) 1 : 4
Q5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Ans. It is connected in parallel.
Q6. A copper wire, has diameter 0.5 mm and resistivity of 1.6 × 10V8 Ω m. What will be the length of this wire to make its resistance 10 Ω ? How much does the resistance change if the diameter is doubled?
Ans. Diameter = 0.5 mm = 0.0005 m
The resistance becomes one-fourth if the diameter is doubled.
Q7. The values of current I flowing in a circuit with resistor for the corresponding values of potential difference V across the resistors are given below:
I (amperes) 0.5 1.0 2.0 3.0 4.0
V(volts) 1.6 3.4 6.7 10.2 13.2
Plot a graph between V and I and calculate the resistance of that resistor.
Ans.
Resistance of resistor R (mean)
Q8. When a 12 V battery is connected across an unknown resistor, there is a current 2.5 inA in the circuit. Find the value of the resistance of the resistor.
Ans. V = 12 V
I = 2.5 mA = 2.5 x 10-3 A