Phy2049 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 = iV = i2R = v2/R

In the diagram R1 > R2 > R3 In the diagram R1 > R2 > R3. 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 4A I. In the diagram, the current in the 3-resistor is 4A. The potential difference between points 1 and 2 is: A. 0.75V B. 0.8V C. 1.25V D. 12V E. 20V

A. 5 Ω B. 7 Ω C. 2 Ω D. 11 Ω E. NOT

(27 – 5)

(27 – 11)

R i motion R i motion motion + - motion + - (27 – 6)

Size and direction of current in E1and E2 Va-Vb R1 = 1 and R2 = 2 Ὠ E1 = 2, E2 = E3 = 4 V Size and direction of current in E1and E2 Va-Vb Kirchhof’s rules: Choose current (magnitude and direction) in all branches. At junction a (or b), I1 = I2+I3 Consider the right box loop, go clockwise. I3(2R1)-E3+E2-I2R2 = 0  I3 = I 2 for the numbers above. And I1 = 2I2 Now the left loop, clockwise: 2 + 2I1 +2I2 - 4 = 0  I2 = 0.33 A = I3 I1 = 0.67 A The current directions are correct as chosen at the outset. . (a) We note that the R1 resistors occur in series pairs, contributing net resistance 2R1 in each branch where they appear. Since e2 = e3 and R2 = 2R1, from symmetry we know that the currents through e2 and e3 are the same: i2 = i3 = i. Therefore, the current through e1 is i1 = 2i. Then from Vb – Va = e2 – iR2 = e1 + (2R1)(2i) we get Therefore, the current through e1 is i1 = 2i = 0.67 A. (b) The direction of i1 is downward. (c) The current through e2 is i2 = 0.33 A. (d) The direction of i2 is upward. (e) From part (a), we have i3 = i2 = 0.33 A. (f) The direction of i3 is also upward. (g) Va – Vb = –iR2 + e2 = –(0.333 A)(2.0 W) + 4.0 V = 3.3 V.

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+4-4-4I=0 I = 2A

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

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

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 -10 +10+10-10+10-10-10-10-10-10 +10 I = 0  I = 4A

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