A `6 V - 12 W` lamp is connected in series with a source of `12 V` supply. Calculate the value of the resistance `R` for the proper working of the lamp. What is the current flowing through the circuit ?

Explanation

Here, power of the lamp, `P = 12 W`
voltage at which it works, `V= 6 V`
As `P = V,I, I = (P)/(V) = (12 W)/(6 V) = 2 A`
Resistance of the lamp, `r = (V)/(I) = (6 V)/(2 A) = 3 Omega`
If this lamp is connected to a source of `12 V` supply directly, current through the lamp `= 12 V//3 Omega = 4 A`.
This current is double the current that the lamp can withstand. Thus, the purpose of the resistance (R) is to reduce the current to the safe value of `2 A`.
Total resistance of the bulb and the resistance `R` (in series) `= r + R`.
For the proper working of the bulb, the resistance `R` should be such a value that `(r + R)` should allow only a current of `2 A` when connected in series with the `12 V` supply, i.e.,
`I = (12 V)/(r + R)` or `2 A = (12 V)/(3 Omega + R)`
or `6 Omega + 2 R = 12 Omega` or `2 R = 6 Omega` or `R = 3 Omega`.