Calculating the Current in a Single-Loop Circuit — Question 4

Question 4

A power ratio identifies two unknown series resistors

Two unknown positive resistors are in series with an ideal 18.0 V source. The common current is 2.00 A and the first resistor dissipates twice the power of the second.

1. Find both resistance values.

2. Find each voltage drop and check the loop and power balances.

Question 4 - Solution

A power ratio identifies two unknown series resistors

(a) Method and calculation

$$R_1+R_2=E/I=9.00\,\Omega,\quad P_1/P_2=I^2R_1/(I^2R_2)=2.$$$$R_1=6.00\,\Omega,\quad R_2=3.00\,\Omega.$$

(b) Interpretation and further calculation

$$V_1=IR_1=12.0\,\mathrm V,\quad V_2=6.00\,\mathrm V,\quad V_1+V_2=18.0\,\mathrm V.$$$$P_1+P_2=24.0+12.0=36.0\,\mathrm W=EI.$$

Physical check

In series, power is proportional to resistance because current is common to both elements.