Ratio of resistances of two bulbs is 2:3. If they are connected in series to a supply, then the ratio of voltages across them is with explanation pls

Explanation:

Given that,

Two resistor has resistance in the ratio 2:3

Then,

R1 : R2 = 2:3

R1 / R2 =⅔

3 •R1 = 2• R2

Let R2 = R

Then,

R1 = ⅔R2 = 2/3 R

So, if the resistor are connected in series

Let know the current that will flow in the circuit

Series connection will have a equivalent resistance of

Req = R1 + R2

Req = R + ⅔ R = 5/3 R

Req = 5R / 3

Let a voltage V be connect across then, the current that flows can be calculated using ohms law

V = iR

I = V/Req

I = V / (5R /3)

I = 3V / 5R

This the current that flows in the two resistors since the same current flows in series connection

Now, using ohms law again to calculated voltage in each resistor

V= iR

For R1 = ⅔R

V1 =i•R1

V1 = 3V / 5R × 2R / 3

V1 = 3V × 2R / 5R × 3

V1 = 2V / 5

For R2 = R

V2 = i•R2

V2 = 3V / 5R × R

V2 = 3V × R / 5R

V2 = 3V / 5

Then,

Ratio of voltage 1 to voltage 2

V1 : V2 = V1 / V2 = 2V / 5 ÷ 3V / 5

V1 : V2 = 2V / 5 × 5 / 3V.

V1 : V2 =2 / 3

V1:V2 = 2:3

The ratio of their voltages is also 2:3

the answer is b, have negligible mass

your answer is c! hope this : )