A circuit with a resistance of #3 Omega# has a fuse with a capacity of #4 A#. Can a voltage of #2 V# be applied to the circuit without blowing the fuse?

1 Answer
Jan 23, 2016

Answer:

The current in a circuit where a voltage of #2# #V# passes through a resistance of #3# #Omega# is given by: #I=V/R = 2/3# #A#. This is well short of the capacity of the fuse, so the fuse will not blow.

Explanation:

Ohm's Law relates voltage #V# #(V)#, current #I# #(A)# and resistance #R# #(Omega)#:

#V=IR#

In this case we want to know the current, so we rearrange to make #I# the subject:

#I=V/R = 2/3# #A#

The current flowing in the circuit is #2/3# #A#. A fuse is designed to 'blow' (burn out) if the current in the circuit is more than its rated value, in this case #4# #A#. This is to protect the rest of the circuit from excessive current which generates heat.

The current in this circuit, #2/3# #A#, is considerably less than #4# #A#, so the fuse will not blow.