A 10 amp fuse is connected in series to 4 identical 5 ohm resistors in parallel with each other. What is the maximum size battery that could be connected without blowing the fuse?
Rt = 5/4 = 1.25 Ohms.
Vmax = I*Rt = 10 * 1.25 = 12.5 Volts.
Thanks Henry
To determine the maximum size battery that could be connected without blowing the fuse, we need to calculate the total current flowing through the circuit and then compare it to the rating of the fuse.
First, let's calculate the equivalent resistance of the four 5-ohm resistors in parallel. When resistors are connected in parallel, the total resistance is given by the formula:
1/Req = 1/R1 + 1/R2 + 1/R3 +...
Since all the resistors are identical and have the same value of 5 ohms, we have:
1/Req = 1/5 + 1/5 + 1/5 + 1/5
= 4/5
Taking the reciprocal of both sides gives:
Req = 5/4 ohms
Now, let's calculate the total current flowing through the circuit. We know that the total current (Itotal) is equal to the current passing through the fuse (Ifuse) since they are connected in series. According to Ohm's law, we have:
Itotal = Vtotal / Req
where Vtotal is the total voltage across the resistors.
Since the fuse has a rating of 10 amps, the total current must not exceed this value. So, we have:
Itotal ≤ 10 amps
Rearranging the earlier formula, we can calculate the maximum voltage allowed:
Vtotal ≤ Itotal * Req
Substituting the values we obtained earlier:
Vtotal ≤ 10 amps * (5/4 ohms)
Vtotal ≤ 12.5 volts
Therefore, the maximum size battery that could be connected without blowing the fuse is 12.5 volts.