1.

Section-A1. (i) Powers of two electric bulb are 100 W and 200 WBoth are connected from mains of 220V. The ratio of resis-0tance of their filaments will be:10(a) 1:2(b) 2:1(a) 1.d) 4.1explain with solution​

Answer»

Given: P₁=100W, P₂=200W, and V=220V

Answer:

Power=Current×Potenial difference

P=VI......(1)

V=IR (Ohm's law)

I=V/R....(2)

Substituting (2) in (1);

P=V²/R

=>R=V²/P

Risistance for 1ST bulb;

R₁=(220)²/100

R₁=48400/100

Thus, R=484 Ohm......(3)

Now, RESISTANCE for 2nd bulb;

R₂=(220)²/200

and, R=242 Ohm......(4)

Ratio of R₁and R₂:

R₁:R₂=484/242

Therefore, R¹:R²=2:1

Thus, OPTION (b) is the answer

I hope it helps ^_^

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