A 20 volt battery has a 4 ohm internal resistance. There is a 5 amp fuse in the main line of the battery. How many 6 ohm resistors can be connected in parallel across the battery before thefuse blows?

This is a trick question. If the external load is x ohms, then the fraction of the voltage drop across the load will be x/(x+4), or 20x/(x+4) volts.

You can see that if the external load is 0, then the current is 5 amps. But, that would take an infinite number of resistors in parallel.

Summing up:

I= 5 A
V=20 V

From V=IR, we have
R=V/I=20/5=4 Ω
So without anything in parallel with the battery, it is at the limit of tripping the fuse.
Any number of resistors will reduce the main current below 5A.

I bet it was 0.4 ohm internal resistance.

Can someone break this down a little better? Im so confused

To determine the number of 6-ohm resistors that can be connected in parallel across the battery before the fuse blows, we need to calculate the total current drawn by the resistors and compare it to the rating of the fuse.

1. Start by calculating the total resistance of the parallel resistors. In a parallel circuit, the total resistance (Rp) can be calculated as the reciprocal of the sum of the reciprocals of individual resistances (R1, R2, R3,...):

1/Rp = 1/R1 + 1/R2 + 1/R3 + ...

Therefore, for n resistors with resistance R, the formula becomes:
1/Rp = n / R
Rp = R / n

2. Given that each resistor has a resistance of 6 ohms, we can now calculate the total resistance after connecting n resistors in parallel:

Rp = 6 / n

3. Now, let's calculate the total current drawn by the parallel resistors. According to Ohm's Law, the current (I) can be calculated as the voltage (V) divided by the total resistance (Rp):

I = V / Rp

In this case, the voltage is 20 volts:

I = 20 / Rp

4. Next, we need to compare this current to the rating of the fuse, which is 5 amps. If the calculated current is equal to or less than the fuse rating, the circuit is within the limits and the fuse will not blow, but if the current exceeds 5 amps, the fuse will blow.

20 / Rp <= 5

5. Substituting the value of Rp from step 2 into the inequality, we get:

20 / (6 / n) <= 5

6. Simplifying the inequality:

20 * (n / 6) <= 5

20n <= 30

7. Divide both sides of the inequality by 20:

n <= 30 / 20

n <= 1.5

8. Since the number of resistors must be a whole number, we round down the result:

n = 1

Therefore, you can connect at most one 6-ohm resistor in parallel across the battery before the fuse blows.