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

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

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

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

3 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:

4 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.

5 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.

6 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.

7 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.

8 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.

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

10 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  )

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


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