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Chapter 28

Direct Current Circuits (Cont.)

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Outline

Kirchhoff’s rules (28.3) RC circuits (28.4) Electrical meters (28.5)

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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: Iin = Iout

2. Loop rule. The sum of the potential differences across all elements around any closed circuit loop must be zero:

loopclosed

0V

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Sign conventions Note the following sign conventions

when using the 2nd 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.

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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.

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Example 28.8 A multi-loop circuit

Find the currents I1, I2, and I3 in the circuit shown in the figure. (Answer: I1= 2.0 A, I2= -3.0 A, I3= -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.

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Charging a capacitor

t < 0 t > 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.

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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.

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Discharging a capacitor

q(t) = Qe-t/RC

I(t) = dq/dt = -(Q/RC)e-t/RC

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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 )

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Electrical meters

Operation principle of a galvanometer

Ammeter

Voltmeter