Calculating the Current in a Single-Loop Circuit — Question 1

Question 1

One loop current satisfies every series voltage drop

A 26.0 V source with internal resistance 1.00 ohm drives three series resistors of 2.00, 4.00 and 6.00 ohm. Assume steady DC and ideal connecting wires.

1. Write the loop equation and find the common current.

2. Find each external resistor power and verify that source power equals all resistive losses.

Question 1 - Solution

One loop current satisfies every series voltage drop

(a) Method and calculation

$$\mathcal E-I(r+R_1+R_2+R_3)=0.$$$$I=26.0/(1+2+4+6)=2.00\,\mathrm A.$$

(b) Interpretation and further calculation

$$P_1=8.00\,\mathrm W,\quad P_2=16.0\,\mathrm W,\quad P_3=24.0\,\mathrm W.$$$$\mathcal EI=52.0\,\mathrm W=8+16+24+I^2r=48+4.$$

Physical check

Charge is not used up as it passes each resistor; the same current transports energy through every series element.