20 HMK #5 Chp 26c

# 20 HMK #5 Chp 26c - DIRECT-CURRENT CIRCUITS 26.15 26 PART 2...

This preview shows pages 1–3. Sign up to view the full content.

D IRECT -C URRENT C IRCUITS PART 2 26.15. IDENTIFY: In both circuits, with and without R 4 , replace series and parallel combinations of resistors by their equivalents. Calculate the currents and voltages in the equivalent circuit and infer from this the currents and voltages in the original circuit. Use to calculate the power dissipated in each bulb. (a) SET UP: The circuit is sketched in Figure 26.13a. EXECUTE: are in parallel, so their equivalent resistance is given by Figure 26.13a The equivalent circuit is drawn in Figure 26.13b. Figure 26.13b Then For resistors in parallel the voltages are equal and are the same as the voltage across the equivalent resistor, so EVALUATE: Note that which is For resistors in parallel the currents add and their sum is the current through the equivalent resistor. (b) SET UP: EXECUTE: which rounds to 1.12 W. glows brightest. EVALUATE: Note that This equals the power dissipated in the equivalent resistor. (c) SET UP: With removed the circuit becomes the circuit in Figure 26.13c. 26

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
EXECUTE: are in parallel and their equivalent resistance is given by and Figure 26.13c The equivalent circuit is shown in Figure 26.13d. Figure 26.13d (d) SET UP: EXECUTE: (e) EVALUATE: When R 4 is removed, P 1 decreases and P 2 and P 3 increase. Bulb R 1 glows less brightly and bulbs R 2 and R 3 glow more brightly. When R 4 is removed the equivalent resistance of the circuit increases and the current through R 1 decreases. But in the parallel combination this current divides into two equal currents rather than three, so the currents through R 2 and R 3 increase. Can also see this by noting that with R 4 removed and less current through R 1 the voltage drop across R 1 is less so the voltage drop across R 2 and across R 3 must become larger. 26.26. IDENTIFY: Apply the loop rule and junction rule. SET UP: The circuit diagram is given in Figure 26.22. The junction rule has been used to find the magnitude and direction of the current in the middle branch of the circuit. There are no remaining unknown currents. EXECUTE:
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

20 HMK #5 Chp 26c - DIRECT-CURRENT CIRCUITS 26.15 26 PART 2...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online