CBSE Class 8 Answered
Four materials W,X,Y,Z are connected in the electrical circuit shown below.
Asked by m.nilu | 21 Oct, 2018, 02:54: PM
Expert Answer
In the list of materials given in different combination, some of them are conductor of electricity and some of them are insulators .
Let us mark the insulating materials as open and find out which combination of materials satisfy
the required condition i.e., only two bulbs light up.
list of materials given in Combination (i), their type and the circuit is given below
W - Brass - conductor, X - Glass - insulator, Y - porcelain - insulator, Z - Copper - conductor
Since material X and Y are insulators they are shown as open, where as W and Z are conducting.
From the circuit, we see that three bulbs A, B and D light up.
list of materials given in Combination (ii), their type and the circuit is given below
W - wood - insulator , X - plastic - insulator, Y - rubber - insulator, Z - gold - conductor
We see in the circuit, W, X and Y are open beacuse they are insulators.
Circuit is not completed with none of the bulb. Hence no bulb light up
list of materials given in Combination (iii), their type and the circuit is given below
W - iron - conductor, X - glass - insulator , Y - wood - insulator , steel - conductor
Since material X and Y are insulators they are shown as open, where as W and Z are conducting.
From the circuit, we see that three bulbs A, B and D light up.
list of materials given in Combination (iv), their type and the circuit is given below
W - lead - conductor, X - porcelain - insulator , Y - iron - conductor , glass - insulator
Since material X and Z are insulators they are shown as open, where as W and Y are conducting.
From the circuit, we see thatonly two bulbs A and B light up.
Conclusion:- Only for combination (iv) correctly represents the materials W, X, Y and Z in the circuit, so that only two bulbs light up.
Answered by Thiyagarajan K | 23 Oct, 2018, 08:34: AM
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