Answer :
The resistance of light bulb ‘a’ will be four times the resistance of the bulb ‘b’ .
As the power dissipated by the both bulbs are equal
P(a) = P(b) and
The voltage across light bulb a is twice that of light bulb b i.e.
V(a) = 2V(b)
resistance of light bulb b is r
R(b) = r
As per the mathematical expression of power ,
P = V²/R
In case of bulb A ,
P(a) = (V(a))²/ R(a)
In case of bulb B ,
P(b) = (V(b))²/ R(b)
As per the statement, both powers are equal so we can also equate their values
(V(a))²/ R(a) = (V(b))²/ R(b) ……………………. (1)
Substituting the values of the variables in (1), we get
4(V(a))²/ R(a) = (V(b))²/ r
R(a) = 4 r
So, the value of resistance of the bulb ‘a’ is four times the value of the resistance of the bulb ‘b’ .
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