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Parallel Circuit Samples

5) The source voltage in Figure 6-67 is 100V. How much voltage does each of the meters read?

Each meter will read 100V because the resistors are in parallel.

11) In the circuit of Figure 6-69, determine the resistance R2, R3, R4

Es

R1

47ohm

R2 R3 R4

5.03mA

1mA 2.14mA

470uA

Figure 6-69

HINT:

It is sometimes helpful to build a chart listing all the circuit givens in a problem. It is easier to see what information that is available to help solve the circuit’s unknowns. Do this starting out and it will help you see what to look for so you can use Ohm’s and Watt’s Laws to solve the given unknowns.

Resistance

R

T

=

Component

Es

R1

R2

Voltage Current

5.03mA

1mA

2.14mA

47Ω

Power

R3

R4

470µA

We need to determine Es for the parallel combination of Figure 6-69. Looking at the stated values in the problem you should see that the values of R

1 we can use Ohm’s Law to solve for V

R1 because they are in parallel.

Es

=

V

R 1

=

I

R 1

×

R

1 and I

R1

are given in this problem, so

, which is the same voltage as Es, V

R2

, V

R3

, and V

R4

VR 1 1 mA 47 47 mV

Now add these values to the chart as more knowns.

Component

Es

R1

R2

Voltage

47mV

47mV

47mV

Current

5.03mA

1mA

2.14mA

Resistance

R

T

=

47Ω

Power

R3

R4

47mV

47mV

470µA

It’s now easy to see how to use Ohm’s Law to solve for the rest of the problems unknowns.

R

2

=

V

I

R 2

R 2

47 mV

2.14

mA

21.96

I

R

3

I

R 4

=

I

T

I

R 1

I

R 2

I

R 3

R 4

=

V

I

R 3

R 3

=

5.03

mA

47

470

− mV

1

µ

A mA

100

2.14

mA

470

µ

A

1.42

mA

R

4

=

V

I

R 4

R 4

47 mV

1.42

mA

33.10

Don’t forget to add these values to the chart.

15) Find the total resistance for each of the following groups of parallel resistors:

1

=

1

+

1

R R R

T 1 2

R

T

=

R R

2

R

1

+

R

2

(a)

(b)

560Ω, 1kΩ = 358.97Ω

47Ω, 56Ω = 25.55Ω

(c) 1.5kΩ, 2.2kΩ, 10kΩ = 818.86Ω

(d) 1MΩ, 470kΩ, 1kΩ, 2.7MΩ = 996.51 Ω

20) If the total resistance in Figure 6-71 is 389.2Ω, what is the value or R

2

?

R1

680ohm

R2

1

=

1

1

R 2 RT R 1

Figure 6-71

R

1

=

1

1

2 389.2

680

R 2

=

910.1

1.098785

mS

21 What is the total resistance between point A and ground in Figure 6-72 for the following conditions?

SW1 SW2

R1

510kohm

R2

470kohm

R3

910kohm

(a) SW1 and SW2 open = 510kΩ

(b) SW1 closed, SW2 open = 244.59kΩ

(c) SW1 open, SW2 closed = 510kΩ

(d) SW1 and SW2 closed = 192.78kΩ

Figure 6-72

(a) R1 will be the only resistor seen by the voltage source.

(b) R1 and R2 will be connected in parallel as the load seen by the voltage source a.

Use the formula

1

𝑅𝑇

=

1

𝑅1

+

1

𝑅2

to solve for the total resistance

(c) R1, R2, and R3 will be connected in parallel as the load seen by the voltage sources a.

Use the formula

1

𝑅𝑇

=

1

𝑅1

+

1

𝑅2

+

1

𝑅3

to solve for the total resistance

Sample Problems

22) What is the total current in each circuit of Figure 6-73?

Es

10V

R1

33ohm

R2

33ohm

R3

27ohm

Es

25V

R1

1kohm

R2

4.7kohm

R3

560ohm

Figure 6-73(a)

R

T

I

T

=

10.24

=

Es

R

T

10 V

10.24

976.56

mA

Figure 6-73(b)

R

T

=

333.50

I

T

=

Es

R

T

25 V

333.50

74.96

mA

26) Find the values of the unspecified labeled quantities in each circuit of Figure 6-74.

500mA

R3

150mA

10V

100V

R1

1kohm

R2

680ohm

R1 R2

IR2 = ?

I1 I2 I3

100mA

I

R 1

=

I

T

( I

R 2

+

I

Figure 6-74(b)

R 3

) I

R 2

I

R 2

Figure 6-74(a)

=

I

T

I

R 1

=

150 mA

100 mA

50 mA

R

1

=

10 V

100 mA

=

100

R

2

=

10 V

50 mA

=

200

I

R 2

=

V

R 2

R

2

100 V

680

=

147.06

mA

I

R 3

I

R 1

=

V

R 3

R

3

=

500 mA

100

1 k

V

=

100

(147.06

mA mA

+

100 mA )

=

252.94

mA

R

1

=

100 V

=

395.35

252.94

mA

Using Current Division Rule

I

R 3

=

V

R 3

=

100 mA

R

3

I

R 2

=

R

3

×

I

R 3

R

2

1 k

680

×

100 mA

=

147.06

mA

I

R 1

= found as above

29) Find the values of the unspecified quantities in Figure 6-77 on page 219 of the textbook.

I

T

250mA

50m

A

I

2

I

3

Figure 6-77

I

R 2

+

I

R 3

=

250 mA

I

R 3

I

R 2

=

V

R 3

R

3

=

250 mA

100

1.2

− k

V

=

83.33

83.33

mA

= mA

166.67

I

T

=

I

R 1

+

I

R 2

+

I

R 3

300 mA mA

R

1

=

100 V

50 mA

=

2 k

R

2

=

100 V

=

599.99

166.67

mA

30) Determine the current through R

L

in each of the circuits of Figure 6-78.

(a) 3mA (I

1

(b) 10μA (I

1

+ I

2

– I

2

= I

L

= I

)

L

)

(c) 0.5A = 500mA (I

1

+ I

3

– I

2

= I

L

38) Five parallel resistors each handle 250mW. What is the total power?

P

T

=5x250mW = 1.25W

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