# Dr. Jie ZouPHY 13611 Chapter 28 Direct Current Circuits (Cont.)

## Presentation on theme: "Dr. Jie ZouPHY 13611 Chapter 28 Direct Current Circuits (Cont.)"— Presentation transcript:

Dr. Jie ZouPHY 13611 Chapter 28 Direct Current Circuits (Cont.)

Dr. Jie ZouPHY 13612 Outline Kirchhoff’s rules (28.3) RC circuits (28.4) Electrical meters (28.5)

Dr. Jie ZouPHY 13613 Kirchhoff’s rules Kirchhoff’s rules are used to simplify the procedure for analyzing more complex circuits: 1. Junction rule. The sum of the currents entering any junction in a circuit must equal the sum of the currents leaving that junction:  I in =  I out 2. Loop rule. The sum of the potential differences across all elements around any closed circuit loop must be zero:

Dr. Jie ZouPHY 13614 Sign conventions Note the following sign conventions when using the 2 nd Kirchhoff’s rule: If a resistor is traversed in the direction of the current, the potential difference  V across the resistor is –IR (Fig. a). If a resistor is traversed in the direction opposite the current, the potential difference  V across the resistor is +IR (Fig. b). If a source of emf is traversed in the direction of the emf (from – to +), the potential difference  V is +  (Fig. c). If a source of emf is traversed in the direction opposite the emf (from + to -), the potential difference  V is -  (Fig. d). Note: We have assumed that battery has no internal resistance. Each circuit element is assumed to be traversed from left to right.

Dr. Jie ZouPHY 13615 Example 28.7 A single-loop circuit A single-loop circuit contains two resistors and two batteries, as shown in the figure below. (A) Find the current in the circuit. (Answer: -0.33 A) (B) What power is delivered to each resistor? What power is delivered by the 12-V battery? (Answer: 0.87 W, 1.1 W, 4.0 W) Note: The “-” sign for I indicates that the direction of the current is opposite the assumed direction. See P. 879 on the textbook for the Problem-solving hints.

Dr. Jie ZouPHY 13616 Example 28.8 A multi-loop circuit Find the currents I 1, I 2, and I 3 in the circuit shown in the figure. (Answer: I 1 = 2.0 A, I 2 = -3.0 A, I 3 = -1.0 A) Problem-solving skills: Choose a direction for the current in each branch. Apply the junction rule. Choose a direction (clockwise or counterclockwise) to transverse each loop. Apply the loop rule. In order to solve a particular circuit problem, the number of independent equations you need to obtain from the two rules equals the number of unknown currents.

Dr. Jie ZouPHY 13617 Charging a capacitor t < 0t > 0 q(t) = C  (1 – e -t/RC ) = Q (1 – e -t/RC ) I(t) = dq/dt = (  /R)e -t/RC  = RC: time constant of the circuit.

Dr. Jie ZouPHY 13618 Examples Quick Quiz: Consider the circuit in the figure and assume that the battery has no internal resistance. Just after the switch is closed, the current in the battery is (a) zero (b)  /2R (c) 2  /R (d)  /R (e) impossible to determine. After a very long time, the current in the battery is (f) zero, (g)  /2R (h) 2  /R (i)  /R (j) impossible to determine.

Dr. Jie ZouPHY 13619 Discharging a capacitor q(t) = Qe -t/RC I(t) = dq/dt = -(Q/RC)e -t/RC

Dr. Jie ZouPHY 136110 Example 28.12 Discharging a capacitor in an RC circuit Consider a capacitor of capacitance C that is being discharged through a resistor of resistance R. After how many time constants is the charge on the capacitor one-fourth its initial value? (answer: 1.39  )

Dr. Jie ZouPHY 136111 Electrical meters Operation principle of a galvanometer Ammeter Voltmeter

Download ppt "Dr. Jie ZouPHY 13611 Chapter 28 Direct Current Circuits (Cont.)"

Similar presentations