The filament of the electric lamp burns out at a current of 200 mA, the electrical resistance

The filament of the electric lamp burns out at a current of 200 mA, the electrical resistance of the heated filament of the lamp is 35 ohms. What is the maximum allowable voltage on the lamp filament.

I = 200 mA = 0.2 A.

R = 35 ohms.

U -?

Let’s write Ohm’s law for the section of the circuit to which the light bulb is connected: I = U / R, where I is the current in the light bulb at which it burns out, U is the maximum allowable voltage in the light bulb, R is the resistance of the light bulb.

Let us express the maximum allowable voltage in the light bulb: U = I * R.

U = 0.2 A * 35 Ohm = 7 V.

Answer: the maximum allowable voltage in a light bulb is U = 7 V.