Phy 2049 Circuits have capacitors and resistors Capacitors

Phy 2049 • Circuits have capacitors and resistors. • Capacitors in parallel in series • Resistors in parallel in Series • Kirchhof’s two rules: voltage around the loop and currents at a junction. • P = i. V = i 2 R = v 2/R

In the diagram R 1 > R 2 > R 3. Rank the three resistors according to the current in them, least to greatest. A. 1, 2, 3 B. 3, 2, 1 C. 1, 3, 2 D. 3, 1, 3 E. All are the same

I. In the diagram, the current in the 3 -resistor is 4 A. The potential difference between points 1 and 2 is: A. 0. 75 V B. 0. 8 V C. 1. 25 V D. 12 V E. 20 V

A. 5 Ω B. 7 Ω C. 2 Ω D. 11 Ω E. NOT

(27 – 11)

R i motion - + motion (27 – 6)

R 1 = 1 and R 2 = 2 Ὠ E 1 = 2, E 2 = E 3 = 4 V Size and direction of current in E 1 and E 2 Va-Vb Kirchhof’s rules: Choose current (magnitude and direction) in all branches. At junction a (or b), I 1 = I 2+I 3. (a) We note that the R 1 resistors occur in series pairs, contributing net resistance 2 R 1 in each branch where they appear. Since e 2 = e 3 and R 2 = 2 R 1, from symmetry we know that the currents through e 2 and e 3 are the same: i 2 = i 3 = i. Therefore, the current through e 1 is i 1 = 2 i. Then from Vb – Va = e 2 – i. R 2 = e 1 + (2 R 1)(2 i) we get Consider the right box loop, go clockwise. I 3(2 R 1)-E 3+E 2 -I 2 R 2 = 0 I 3 = I 2 for the numbers above. And I 1 = 2 I 2 Now the left loop, clockwise: Therefore, the current through e 1 is i 1 = 2 i = 0. 67 A. (b) The direction of i 1 is downward. (c) The current through e 2 is i 2 = 0. 33 A. (d) The direction of i 2 is upward. (e) From part (a), we have i 3 = i 2 = 0. 33 A. (f) The direction of i 3 is also upward. (g) Va – Vb = –i. R 2 + e 2 = –(0. 333 A)(2. 0 W) + 4. 0 V = 3. 3 V. 2 + 2 I 1 +2 I 2 - 4 = 0 I 2 = 0. 33 A = I 3 I 1 = 0. 67 A The current directions are correct as chosen at the outset.

R = 4 Ω, V = 4 V, what is the current through R?

R = 4 Ω, V = 4 V, what is the current through R? 4+4+4 -4 -4 I=0 I = 2 A

Obtain current in all branches of this circuit by repeated application of Kirchhof’s loop and junction rules

• What is i? All V = 10 Volts and all R = 4 Ohms

j k • What is i? All V = 10 Volts and all R = 4 Ohms? • What is j? What is k?

j k • What is i? All V = 10 Volts and all R = 4 Ohms -10 +10+10 -10 -10 -10 +10 I = 0 I = 4 A

• Three circuits are connected to a battery. Rank them by the – final charge – Time constant