1. 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

2. 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

3. 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

4. A. 5 Ω B. 7 Ω C. 2 Ω D. 11 Ω E. NOT

5. (27 – 5)

6. (27 – 11)

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

8. 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) V = 3.3 V.

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

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

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

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

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

14. What is i? All V = 10 Volts and all R = 4 Ohmsj k
What is i? All V = 10 Volts and all R = 4 Ohms
I = 0  I = 4A

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