## Chapter 19. Electric Circuits

Chapter 19 Electric Circuits Kirchhoff’s Rules How can one deal with a complicated circuit like this? à Use Kirchhoff’s rules: * Junction rule * ...
Author: Ralph Patterson
Chapter 19 Electric Circuits

Kirchhoff’s Rules

How can one deal with a complicated circuit like this?

à Use Kirchhoff’s rules: * Junction rule * Loop rule

Kirchhoff’s Rules

The junction rule states that the total current directed into a junction must equal the total current directed out of the junction.

Kirchhoff’s Rules

The loop rule expresses conservation of energy in terms of the electric potential and states that for a closed circuit loop, the total of all potential rises is the same as the total of all potential drops.

Kirchhoff’s Rules

KIRCHHOFF’S RULES Junction rule. The sum of the magnitudes of the currents directed into a junction equals the sum of the magnitudes of the currents directed out of a junction. Loop rule. Around any closed circuit loop, the sum of the potential drops equals the sum of the potential rises, i.e. the sum over all of the changes in potential is zero.

Kirchhoff’s Rules

Example: Using Kirchhoff’s Loop Rule Determine the current in the circuit.

Kirchhoff’s Rules

Loop around circuit in clockwise direction.

24 V − I (12 Ω) − 6.0 V − I (8.0 Ω) = 0 I (12 Ω) + 6.0 V + I (8.0 Ω) = 24 V   potential rises potential drops

I ( 20 Ω) = 18 V

I = 0.90 A

Kirchhoff’s Rules

Example: A car’s headlight is connected to the battery which is charged by the alternator as shown in the figure and circuit diagram. Find the currents through the headlight, battery and alternator.

Kirchhoff’s Rules

Reasoning Strategy in applying Kirchhoff’s Rules 1.  Draw the current in each branch of the circuit. Choose any direction. If your choice is incorrect, the value obtained for the current will turn out to be a negative number. 2.  Mark each resistor with a + at one end and a – at the other end in a way that is consistent with your choice for current direction in step 1. For batteries, the signs will be the usual + for higher potential and – for lower potential. 3.  Apply the junction rule and the loop rule to the circuit, obtaining in the process as many independent equations as there are unknown variables. 4. Solve these equations simultaneously for the unknown variables.

rA = 0.100 Ω rB = 0.010 Ω RH = 1.20 Ω Apply the Junction rule at B:

IA + IB = IH

Eq. 1

Apply the Loop rule for ABEF and BCDE In the clockwise direction: ABEF:

BCDE:

2 + I B rB 14 − I A rA + I B rB −12 = 0 ⇒ I A = rA 12 − I B rB 12 − I B rB − I H RH = 0 ⇒ I H = RH

Eq. 2

Eq. 3

Put Eq. 2 and Eq. 3 into Eq. 1, solve for IB. Then use Eq. 2 and Eq. 3 to get IA and IH:

2 + I B rB 12 − I B rB + IB = rA RH ⇒ IB =

12rA − 2RH = −9.02 A I A = 19.10 A I H = 10.08 A rB RH + rA RH + rA rB

IB flows in opposite direction of diagram

Capacitors in Series and Parallel

q = q1 + q2 = C1V + C2V = (C1 + C2 )V

Parallel capacitors

CP = C1 + C2 + C3 + 

Capacitors in Series and Parallel

⎛ 1 q q 1 ⎞ V = V1 + V2 = + = q⎜⎜ + ⎟⎟ C1 C2 ⎝ C1 C2 ⎠

Series capacitors

1 1 1 1 = + + + CS C1 C2 C3

Example: Two capacitors have values 2.7 µF and 4.2 µF. Find their equivalent capacitance if they are connected in a) series, and b) parallel with each other. If the capacitive networks in a) and b) are connected across a 200 V source, find c) the energy stored in them in each case.

a)  Series

1 1 1 5 −1 = + = 6.1×10 F −6 −6 Cs 2.7 ×10 4.2 ×10

b)  Parallel C p = 2.7 ×10 −6 + 4.2 ×10 −6 = 6.9 ×10 −6 F

1 CV 2 2 1 2 −6 Energys = (1.6 ×10 ) ( 200 ) = 0.032 J 2 1 2 −6 Energy p = ( 6.9 ×10 ) ( 200 ) = 0.14 J 2

c)  Energy =

Cs = 1.6 ×10 −6 F

RC Circuits

Capacitor charging

−t RC \$ " q = qo #1− e %

time constant

τ = RC

RC Circuits

Capacitor discharging

q = qo e −t RC

time constant

τ = RC

Example: How long does it take to discharge an RC circuit to the fraction 1/e (about 37%) of its original charge for a 1000 Ω resistor connected in series with a) a 500 µF capacitor, and b) a 0.5 F capacitor?

a) τ = RC = (1000 ) 50 ×10 −6 = 0.05 s = 50 ms

(

)

b) τ = RC = (1000 ) ( 0.5) = 500 s = 8.3 min

The measurement of current and voltage in DC circuits Devices used: Ammeter -- measures current flowing in the circuit Voltmeter -- measures voltage across some device in the circuit Ammeters and voltmeters can be either analog (read out with the deflection of a needle) or digital devices. We will study how the analog devices work since they’re easier to understand from basic principles. Galvanometer -- an analog device that responds to electrical currents flowing through it by causing a needle to deflect across some scale. Both the ammeter and voltmeter are based on a galvanometer.

The Measurement of Current and Voltage

How a galvanometer works. The coil of wire and pointer rotate when there is a current in the wire. This one is calibrated so that if a current of 0.10 mA flows through the galvanometer coil, a full-scale deflection of the needle on the calibrated scale Is obtained.

Schematic representation of a galvanometer showing the resistance in its coil RC in series with the galvanometer.

The Measurement of Current and Voltage

Using the galvanometer as an ammeter. If a galvanometer with a full-scale limit of 0.100 mA is to be used to measure the current of 60.0 mA, a parallel shunt resistance must be used so that the excess current of 59.9 mA can detour around the galvanometer coil. Assuming RC = 50 Ω, find R. VAB = IGRC = (0.1 x 10-3)(50) = 5 x 10-3 V R = VAB/IR = (5 x 10-3)/(59.9 x 10-3) = 8.35 x 10-2 Ω

The Measurement of Current and Voltage

An ammeter must be inserted into a circuit so that the current passes directly through it. Thus, it is important that it has as low a resistance as possible so as to minimize its effect on the circuit since it acts as a series resistor added to the circuit. Find the equivalent resistance of the ammeter in our example to see if it is small. RC and R are in parallel, and RC = 50 Ω and R = 8.35 x 10-2 Ω

1/RA = 1/RC + 1/R = 1/(50) + 1/(8.35 x 10-2) = 12.0 Ω-1 --> RA = 0.083 Ω

Circuits generally have resistances much higher than this, ~102 to ~103 Ω

Using the galvanometer as a voltmeter If a galvanometer with a full-scale limit of 0.100 mA is to be used to measure the voltage of 10.0 V, a series resistor must be used to limit the current flowing through the galvanometer to 0.100 mA. Assuming RC = 50 Ω, find R.

A VAB

RV = R + RC , since R and RC in series

B

VAB = IRV = I(R+RC)

ê

R = VAB/I - RC = (10.0)/(0.100 x 10-3) - 50 = 105 Ω

R

A

VAB

B

The Measurement of Current and Voltage

To measure the voltage between two points in a circuit, a voltmeter is connected between the points. Thus, it is important that it has as high a resistance as possible so as to minimize its effect on the circuit since it acts as a parallel resistor added to the circuit. Find the equivalent resistance of the voltmeter in our example to see if it is large. R and RC are in series, and R = 105 Ω and RC = 50 Ω RV = R + RC = 105 + 50 = 105 Ω Circuits generally have resistances much lower than this, ~102 to ~103 Ω

Safety in grounding electrical devices

To reduce the danger inherent in using circuits, proper electrical grounding is necessary.