Question

A light bulb has a resistance of 8ohms and a maximum current of 10a. How much voltage can be applied before the bulb will break?

Answer 1

YOUR ANSWER IS

NOW

I= 10A

R = 8OOHMS

NOW

V = IR

V = 10X 8 = V = 80VOLT

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Answer 2

Given that, a light bulb has a resistance of 8ohms and a maximum current of 10 A.

We have to find the voltage of the bulb that can be applied before the bulb will break.

From above data we have; Resistance 8 ohm and current 10A.

Ohm's law: The potential difference i.e. V

across the ends of the wire is directly

proportional to the current i.e. I flowing

through that wire provides the temperature

remains constant.

From ohm's law we can say that,

V = IR

Here; R is Resistance, I is current and V is voltage.

Unit of resistance is ohm, current is A (ampere) and Voltage is V (volt).

Substitute the known values in the above formula,

→ V = 8 × 10

→ V = 80

Therefore, the voltage of the bulb is 80V.