Two bulbs are marked 100w, 220 v and 60 w, 110v. calculate the ratio of their resistance.

44.

Three resistors are connected to a 12 V battery as shown in the figure given below:

i) What is the current through the 8 ohm resistor?ii) What is the potential difference across the parallel combination of 6 ohm and 12 ohm resistor?

iii) What is the current through the 6 ohm resistor?

Here, 12 ohm and 6 ohm resistors are connected in parallel.

Therefore, their effective resistance, 

i) Current drawn = 

ii) P.D. across the 8 ohm resistor = I  R = 1  = 8 VP.D. across the parallel combination of resistors = 12 – 8 = 4 V

Because, the resistances are in parallel and they have the same P.D. each.

That is equal to 4 V.

iii) Current through the 6 ohm resistor =

Text Solution

Solution : Given : <br> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/GRU_ICSE_10Y_SP_X_PHY_11_E01_022_S01.png" width="80%"> <br> We know, `P=V^(2)/R` <br> or `R=V^(2)/P` <br> `thereforeR_(1)/R_(2)=((V_(1)^(2))/P_(1))/((V_(2)^(2))/P_(2))` <br> = `(V_(1)^(2)xxP_(2))/(P_(1)xxV_(2)^(2))=((220)^(2)xx60)/(100xx(110)^(2))` <br> = `12/5` <br> `R_(1):R_(2)=12:5`

Two bulbs are marked 100 W, 220 V and 60 W, 110 V. Calculate the ratio of their resistances.

Given Ist Bulb 

P1 = 100 W

V1 = 220 V 

IInd Bulb

P2 = 60 W

V2 = 110V

We know P = `"V"^2/"R"`

So R = `"V"^2/"P"`

So `"R"_1/"R"_2 = ("V"_1^2/"P"_1)/("V"_2^2/"P"_2)`

`= ("V"_1^2 xx "P"_2)/("P"_1 xx "V"_2^2)`

`= (220^2 xx 60)/(100 xx 110 xx 110) = 2.4`

Concept: Electric Potential (Electrostatic Potential) and Potential Difference

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