To solve the problem of combining several identical resistors of \(10 \, \Omega\) each, capable of carrying a maximum current of \(1 \, A\), to produce a resistance of \(5 \, \Omega\) that can carry a total current of \(4 \, A\), we can follow these steps:
### Step 1: Understand the requirements
We need to create a total resistance of \(5 \, \Omega\) and ensure that the combination can handle a total current of \(4 \, A\). Since each resistor can only handle \(1 \, A\), we will need to arrange them in such a way that we can achieve this current capacity.
### Step 2: Determine the configuration
To achieve a total current of \(4 \, A\), we can use parallel branches. Each branch can carry \(1 \, A\), so we will need \(4\) branches to carry a total of \(4 \, A\).
### Step 3: Calculate the equivalent resistance of each branch
Let \(R_s\) be the resistance of each branch. Since we need \(4\) branches in parallel, the total resistance \(R_{net}\) can be expressed as:
\[
\frac{1}{R_{net}} = \frac{1}{R_s} + \frac{1}{R_s} + \frac{1}{R_s} + \frac{1}{R_s} = \frac{4}{R_s}
\]
Thus,
\[
R_{net} = \frac{R_s}{4}
\]
### Step 4: Set up the equation for total resistance
We want \(R_{net} = 5 \, \Omega\). Therefore, we can set up the equation:
\[
5 = \frac{R_s}{4}
\]
From this, we can solve for \(R_s\):
\[
R_s = 5 \times 4 = 20 \, \Omega
\]
### Step 5: Determine the number of resistors in each branch
Since each resistor has a resistance of \(10 \, \Omega\), we need to find out how many resistors are required to make \(R_s = 20 \, \Omega\). If we connect two \(10 \, \Omega\) resistors in series, we get:
\[
R_s = 10 + 10 = 20 \, \Omega
\]
Thus, we need \(2\) resistors in each branch.
### Step 6: Calculate the total number of resistors
Since we have \(4\) branches and each branch requires \(2\) resistors, the total number of resistors required is:
\[
\text{Total resistors} = 4 \times 2 = 8
\]
### Conclusion
The minimum number of resistors required to achieve the desired resistance and current capacity is \(8\).
