Calculating the Current in a Single-Loop Circuit — Question 9

Question 9

Series power ratings constrain the source voltage through one common current

Two series resistors have $R_1=100\,\Omega$, rating .250 W, and $R_2=220\,\Omega$, rating .500 W. They are connected to an ideal variable DC source.

1. Find the largest allowed common current and identify the limiting resistor.

2. Find the largest source voltage and the power in the nonlimiting resistor at that setting.

Question 9 - Solution

Series power ratings constrain the source voltage through one common current

(a) Method and calculation

$$I\le\sqrt{.250/100}=.0500\,\mathrm A,\quad I\le\sqrt{.500/220}=.0476731\,\mathrm A.$$$$I_{\max}=.0476731\,\mathrm A\quad\text{limited by }R_2.$$

(b) Interpretation and further calculation

$$E_{\max}=I_{\max}(100+220)=15.2554\,\mathrm V.$$$$P_1=I_{\max}^2(100)=.227273\,\mathrm W<.250\,\mathrm W.$$

Physical check

The higher nominal power rating does not alone determine which element limits a series current.