Question

A $100\ \Omega$, 1 W resistor and a $800\ \Omega$, 2 W resistor are connected in series. The maximum DC voltage that can be applied continuously to the series circuit without exceeding the power limit of any of the resistors is:

• A. 90 V
• B. 50 V
• C. 45 V
• D. 40 V

Hint:

When multiple resistors are in series with different power ratings, use $P = I^2 R$ to find the limiting current for each, and choose the lower one to ensure safety.

Answer 1

The Correct Option is C

Let the total voltage across the series combination be $V$.
Power dissipated by a resistor in a series circuit is given by:
\[ P = I^2 R \]
Let the total resistance be: \[ R_{\text{total}} = 100 + 800 = 900\ \Omega \]
Let current in the circuit be $I$.
We first ensure that the smaller resistor (100 $\Omega$, 1 W) does not exceed its power rating.
Using $P = I^2 R$ for the 100 $\Omega$ resistor: \[ 1 = I^2 \times 100 ⇒ I^2 = \frac{1}{100} ⇒ I = \sqrt{0.01} = 0.1\ \text{A} \]
Now calculate total voltage across the series circuit: \[ V = I \times R_{\text{total}} = 0.1 \times 900 = 90\ \text{V} \]
Now check power dissipation in the 800 $\Omega$ resistor: \[ P = I^2 \times 800 = (0.1)^2 \times 800 = 0.01 \times 800 = 8\ \text{W} \] This exceeds the 2 W rating! Hence, we must reduce current to ensure both resistors remain within power limit.
Now calculate maximum current that the 800 $\Omega$, 2 W resistor can handle: \[ 2 = I^2 \times 800 ⇒ I^2 = \frac{2}{800} = 0.0025 ⇒ I = \sqrt{0.0025} = 0.05\ \text{A} \]
Now recalculate total voltage using $I = 0.05\ \text{A}$:
\[ V = I \times R_{\text{total}} = 0.05 \times 900 = 45\ \text{V} \]
This value keeps both resistors within their power limits.
Hence, the maximum voltage that can be applied safely is 45 V.