Can somebody please explain me example 12.5 from chapter 12 electricity of ncert book class 10??
Answers
I gave my Boards nearly so I can explain.
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 the ratio R/R ′ is –
(a) 1/25
(b) 1/5
(c) 5
(d) 25
Answer. Correct option is (d)
Let’s now justify this answer
Resistance of a piece of wire is proportional to its length. According to question we have a piece of wire of resistance R. The wire is cut into five equal parts. Therefore resistance of each part would be
Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions
Since Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions
All five resistance are connected in parallel. So, for parallel resistance
Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions
Therefore the ratio of R/R’ is 25.
Question 2. Which of the following terms does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
(d) V2/R
Answer. the rate at which electric work is done or the rate at which electric energy is consumed is called electric power.
Electric power is given by the expression
Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions (i)
According to Ohm’s Law
Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions (ii)
Where,
V = Potential difference
I = Current
R = resistance
Equation (i) can be written as
Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions
Again considering equation (ii)
Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions
Power can not be expressed as IR2.
Question 3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be – (a) 100 W (b) 75 W (c) 50 W (d) 25 W
Answer.(d) Power of an appliance determine the rate at which the electrical energy is delivered to it.
Energy consumed by an appliance is given by th expression
Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions
Where,
Power rating, P=100 W
Voltage, V=220 V
Resistance,
Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions
The resistance of the bulb remains constant if the supply voltage is reduced to 110 V.
If the bulb is operated on 110 V, then the energy consumed by it is given by the expression for power as
Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions
Therefore , the power consumed would be 25 W.
Question 4. 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
Answer.(c) It is given that resistors are equal.
So, for series combination
Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions
Power in series combination is
For parallel combination of resistances
Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions
Power in parallel combination is
Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions
So,
Class 10 Science: Chapter 12 electricity Exercise Questions|NCERT Solutions
Question 5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer. To measure the potential difference between two points, a voltmeter should be connected in parallel to the points. Also its positive terminal should be connected to the point at higher potential and negative terminal to the point at lower potential.
Question 6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 ? m. What will be the length of this wire to make its resistance 10 ?? How much does the resistance change if the diameter is doubled?
Answer. Given that
D=0.5 mm,