A circuit with a resistance of #8 Omega# has a fuse with a capacity of #3 A#. Can a voltage of #36 V# be applied to the circuit without blowing the fuse?

Question

A circuit with a resistance of #8 Omega# has a fuse with a capacity of #3 A#.  Can a voltage of #36 V# be applied to the circuit without blowing the fuse?

Emily Abbate · Accepted Answer

No.

Well, we need to find the voltage that the fuse can hold, so we use Ohm's law, which states that, #V=IR# where: #V# is the voltage in volts #I# is the current in amperes #R# is the resistance in ohms So, we get: #V=3 \ "A"*8 \ Omega# #=24 \ "V"# Since a voltage of #36# volts is applied to the circuit, but the fuse only holds #24# volts, it'll blow up the fuse.

Charlotte Aaron · Answer

undefined Yes, a voltage of 36 V can be applied to the circuit without blowing the fuse. To determine if the fuse will blow, we can use Ohm's law, which states that voltage ($V$) equals current ($I$) multiplied by resistance ($R$):

$$ V = I \times R $$

Rearranging the formula to solve for current ($I$), we have:

$$ I = \frac{V}{R} $$

Substituting the given values:

$$ I = \frac{36 \, \text{V}}{8 \, \Omega} $$

$$ I = 4.5 \, \text{A} $$

Since the current ($I = 4.5 \, \text{A}$) is less than the capacity of the fuse ($3 \, \text{A}$), the voltage of 36 V can be applied to the circuit without blowing the fuse.