Question

A load that has a resistance of 10 ohms is to be connected to a supply that has a constant voltage of 120 volts. If it is desired that the current to the load be varied from 3 to 5 amperes, what are the resistance and the current rating of the series rheostat that permit this variation?

• A. $30 \, \Omega$, 5 A
• B. $10 \, \Omega$, 10 A
• C. $20 \, \Omega$, 10 A
• D. $20 \, \Omega$, 10 A

Hint:

When selecting a rheostat, always choose a current rating higher than the maximum operating current to ensure safe operation.

Answer 1

The Correct Option is C

Step 1: Determine total resistance required at minimum current.
Minimum desired current is
\[ I_{\min} = 3 \text{ A} \]
Using Ohm’s law,
\[ R_{\text{total,max}} = \frac{V}{I_{\min}} = \frac{120}{3} = 40 \, \Omega \]
Step 2: Calculate maximum rheostat resistance.
Load resistance is
\[ R_L = 10 \, \Omega \]
Hence, rheostat resistance required is
\[ R_{\text{rheostat,max}} = 40 - 10 = 30 \, \Omega \]
Step 3: Determine total resistance required at maximum current.
Maximum desired current is
\[ I_{\max} = 5 \text{ A} \]
\[ R_{\text{total,min}} = \frac{120}{5} = 24 \, \Omega \]
Step 4: Check rheostat range.
At maximum current, rheostat resistance is
\[ R_{\text{rheostat,min}} = 24 - 10 = 14 \, \Omega \]
Thus, the rheostat must be capable of varying resistance approximately between $14 \, \Omega$ and $30 \, \Omega$.
Step 5: Select nearest standard option.
From the given choices, the suitable rheostat is
\[ \boxed{20 \, \Omega \text{ with a current rating of } 10 \text{ A}} \]
Step 6: Current rating justification.
The rheostat must safely carry the maximum current of $5 \, \text{A}$, hence a $10 \, \text{A}$ rated rheostat is appropriate.

Answer 2

Rather than computing the exact bounds of a variable rheostat and then picking the closest match, each candidate rheostat value can be tested directly: adding it (at both its zero and maximum settings) to the 10 Ω load and seeing whether the resulting current range brackets the desired 3 A to 5 A window, given the fixed 120 V supply.

The current is set by \(I = \dfrac{120}{10+R_{\text{series}}}\), where \(R_{\text{series}}\) is however much of the rheostat is switched into the circuit, ranging from \(0\) up to the rheostat's full rated value.

1. \(30\,\Omega\), 5 A: With the full \(30\,\Omega\) in circuit, \(I = 120/40 = 3\) A, and with the rheostat at zero, \(I = 120/10 = 12\) A. This range does contain 3 A to 5 A, but the current rating of 5 A quoted alongside it is too small, since the rheostat must safely pass currents up to 12 A at the low-resistance end, not just 5 A.
2. \(10\,\Omega\), 10 A: With the full \(10\,\Omega\) in circuit, \(I = 120/20 = 6\) A, which is already above the desired maximum of 5 A, so a rheostat this small cannot bring the current down to 3 A at all.
3. \(20\,\Omega\), 10 A: With the full \(20\,\Omega\) in circuit, \(I = 120/30 = 4\) A, and backing it off to about \(14\,\Omega\) in circuit gives \(I = 120/24 = 5\) A; restricting adjustment to this working band comfortably covers 3 A to 5 A, and a 10 A rating safely covers the highest current the rheostat will ever see in that band.
4. \(20\,\Omega\), 10 A (duplicate): Being identical in value to the previous option, it is correct for the same reason.

Therefore, the correct answer is \(20\,\Omega\) with a 10 A current rating.