Find the current values i 1 ( 0 − ) and i 2 ( 0 − ) .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 7, Problem 1P

a.

To determine

Find the current values i1(0−) and i2(0−).

a.

Expert Solution

Answer to Problem 1P

The current values i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

Explanation of Solution

PSPICE Circuit:

Refer to the Figure P7.1 in the textbook.

Draw the given circuit diagram in PSPICE as shown in Figure 1.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 2

PSPICE output:

After run the PSPICE circuit a black output screen will be displayed. Right click on the mouse by keeping cursor on the output screen, click the option “Add Trace” and type the expression “-I(R3)” in trace expression box.

The current plot i1(t) is shown in Figure 3.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 3

From PSPICE output, the initial values of current are,

i1(0−)=5 mAi1(0−)=i1(0+)i1(0+)=5 mA

Similarly, type the expression “-I(R2)” in trace expression box to obtain the current i2(t).

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 4

From PSPICE output, the initial values of current are,

i2(0−)=15 mAi2(0+)=−5mA

Conclusion:

Therefore, the values of i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

b.

To determine

Find the current values i1(0+) and i2(0+).

b.

Expert Solution

Answer to Problem 1P

The current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

Explanation of Solution

Calculation:

From Figure 2 and Figure 3, the current values are,

i1(0+)=5 mAi2(0+)=−5mA

Conclusion:

Therefore, the current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

c.

To determine

Find the expression i1(t) for t≥0.

c.

Expert Solution

Answer to Problem 1P

The expression i1(t) for t≥0 is 5e−20,000t mA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i1(t) is,

i1(t)=i1(0+)e−tτ

Substitute 5 mA for i1(0+) and 50 μs for τ.

i1(t)=5e−t50 μs mA=5e−20,000t mA

Conclusion:

Therefore, the expression i1(t) for t≥0 is 5e−20,000t mA.

d.

To determine

Find the expression i2(t) for t≥0.

d.

Expert Solution

Answer to Problem 1P

The expression i2(t) for t≥0 is −5e−20,000t mA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i2(t) is,

i2(t)=i2(0+)e−tτ

Substitute –5 mA for i1(0+) and 50 μs for τ.

i2(t)=−5e−t50 μs mA=−5e−20,000t mA

Conclusion:

Therefore, the expression i2(t) for t≥0 is −5e−20,000t mA.

e.

To determine

Explain the reason for why i2(0−)≠i2(0+).

e.

Expert Solution

Explanation of Solution

Calculation:

The current in the resistor changes instantaneously. The switching operation makes the current i2(0−) is equal to 15 mA and i2(0+) is equal to –5 mA.

Conclusion:

Therefore, the reason for why i2(0−)≠i2(0+) is stated.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Find the valve of the voltage Vx using the THEVENIN equivalent circuit and redo the problem with the NORTON equivalent circuit. Show both the the vinen and Norton circuits. I 12V m 1 ww 3 23 + 43Vx 5 63 ми

Find the valve of V using the Thevenin Equivalent Circuit and then determine if the 8 ohm resistor allows maximum power transfer. If not, then what value should the 8 ohm resistor be changed to for maximum power transfer? ZA 6 6 + 22V 83 V 34 2 6 АААА АААА

Find the valve of voltage Vx using the THE VIN IN equivalent circuit ww 8 Show the Theven in Circuit. I 7V ZV m 6 5 M + 4 34 АА 3 1 АААА 9A ↑ 24

Answer 2

Question

Chapter 7, Problem 1P

a.

To determine

Find the current values i1(0−) and i2(0−).

a.

Expert Solution

Answer to Problem 1P

The current values i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

Explanation of Solution

PSPICE Circuit:

Refer to the Figure P7.1 in the textbook.

Draw the given circuit diagram in PSPICE as shown in Figure 1.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 2

PSPICE output:

After run the PSPICE circuit a black output screen will be displayed. Right click on the mouse by keeping cursor on the output screen, click the option “Add Trace” and type the expression “-I(R3)” in trace expression box.

The current plot i1(t) is shown in Figure 3.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 3

From PSPICE output, the initial values of current are,

i1(0−)=5 mAi1(0−)=i1(0+)i1(0+)=5 mA

Similarly, type the expression “-I(R2)” in trace expression box to obtain the current i2(t).

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 4

From PSPICE output, the initial values of current are,

i2(0−)=15 mAi2(0+)=−5mA

Conclusion:

Therefore, the values of i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

b.

To determine

Find the current values i1(0+) and i2(0+).

b.

Expert Solution

Answer to Problem 1P

The current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

Explanation of Solution

Calculation:

From Figure 2 and Figure 3, the current values are,

i1(0+)=5 mAi2(0+)=−5mA

Conclusion:

Therefore, the current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

c.

To determine

Find the expression i1(t) for t≥0.

c.

Expert Solution

Answer to Problem 1P

The expression i1(t) for t≥0 is 5e−20,000t mA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i1(t) is,

i1(t)=i1(0+)e−tτ

Substitute 5 mA for i1(0+) and 50 μs for τ.

i1(t)=5e−t50 μs mA=5e−20,000t mA

Conclusion:

Therefore, the expression i1(t) for t≥0 is 5e−20,000t mA.

d.

To determine

Find the expression i2(t) for t≥0.

d.

Expert Solution

Answer to Problem 1P

The expression i2(t) for t≥0 is −5e−20,000t mA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i2(t) is,

i2(t)=i2(0+)e−tτ

Substitute –5 mA for i1(0+) and 50 μs for τ.

i2(t)=−5e−t50 μs mA=−5e−20,000t mA

Conclusion:

Therefore, the expression i2(t) for t≥0 is −5e−20,000t mA.

e.

To determine

Explain the reason for why i2(0−)≠i2(0+).

e.

Expert Solution

Explanation of Solution

Calculation:

The current in the resistor changes instantaneously. The switching operation makes the current i2(0−) is equal to 15 mA and i2(0+) is equal to –5 mA.

Conclusion:

Therefore, the reason for why i2(0−)≠i2(0+) is stated.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 7, Problem 1P

a.

To determine

Find the current values i1(0−) and i2(0−).

a.

Expert Solution

Answer to Problem 1P

The current values i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

Explanation of Solution

PSPICE Circuit:

Refer to the Figure P7.1 in the textbook.

Draw the given circuit diagram in PSPICE as shown in Figure 1.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 2

PSPICE output:

After run the PSPICE circuit a black output screen will be displayed. Right click on the mouse by keeping cursor on the output screen, click the option “Add Trace” and type the expression “-I(R3)” in trace expression box.

The current plot i1(t) is shown in Figure 3.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 3

From PSPICE output, the initial values of current are,

i1(0−)=5 mAi1(0−)=i1(0+)i1(0+)=5 mA

Similarly, type the expression “-I(R2)” in trace expression box to obtain the current i2(t).

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 4

From PSPICE output, the initial values of current are,

i2(0−)=15 mAi2(0+)=−5mA

Conclusion:

Therefore, the values of i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

b.

To determine

Find the current values i1(0+) and i2(0+).

b.

Expert Solution

Answer to Problem 1P

The current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

Explanation of Solution

Calculation:

From Figure 2 and Figure 3, the current values are,

i1(0+)=5 mAi2(0+)=−5mA

Conclusion:

Therefore, the current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

c.

To determine

Find the expression i1(t) for t≥0.

c.

Expert Solution

Answer to Problem 1P

The expression i1(t) for t≥0 is 5e−20,000t mA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i1(t) is,

i1(t)=i1(0+)e−tτ

Substitute 5 mA for i1(0+) and 50 μs for τ.

i1(t)=5e−t50 μs mA=5e−20,000t mA

Conclusion:

Therefore, the expression i1(t) for t≥0 is 5e−20,000t mA.

d.

To determine

Find the expression i2(t) for t≥0.

d.

Expert Solution

Answer to Problem 1P

The expression i2(t) for t≥0 is −5e−20,000t mA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i2(t) is,

i2(t)=i2(0+)e−tτ

Substitute –5 mA for i1(0+) and 50 μs for τ.

i2(t)=−5e−t50 μs mA=−5e−20,000t mA

Conclusion:

Therefore, the expression i2(t) for t≥0 is −5e−20,000t mA.

e.

To determine

Explain the reason for why i2(0−)≠i2(0+).

e.

Expert Solution

Explanation of Solution

Calculation:

The current in the resistor changes instantaneously. The switching operation makes the current i2(0−) is equal to 15 mA and i2(0+) is equal to –5 mA.

Conclusion:

Therefore, the reason for why i2(0−)≠i2(0+) is stated.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 7, Problem 1P

a.

To determine

Find the current values i1(0−) and i2(0−).

a.

Expert Solution

Answer to Problem 1P

The current values i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

Explanation of Solution

PSPICE Circuit:

Refer to the Figure P7.1 in the textbook.

Draw the given circuit diagram in PSPICE as shown in Figure 1.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 2

PSPICE output:

After run the PSPICE circuit a black output screen will be displayed. Right click on the mouse by keeping cursor on the output screen, click the option “Add Trace” and type the expression “-I(R3)” in trace expression box.

The current plot i1(t) is shown in Figure 3.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 3

From PSPICE output, the initial values of current are,

i1(0−)=5 mAi1(0−)=i1(0+)i1(0+)=5 mA

Similarly, type the expression “-I(R2)” in trace expression box to obtain the current i2(t).

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 4

From PSPICE output, the initial values of current are,

i2(0−)=15 mAi2(0+)=−5mA

Conclusion:

Therefore, the values of i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

b.

To determine

Find the current values i1(0+) and i2(0+).

b.

Expert Solution

Answer to Problem 1P

The current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

Explanation of Solution

Calculation:

From Figure 2 and Figure 3, the current values are,

i1(0+)=5 mAi2(0+)=−5mA

Conclusion:

Therefore, the current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

c.

To determine

Find the expression i1(t) for t≥0.

c.

Expert Solution

Answer to Problem 1P

The expression i1(t) for t≥0 is 5e−20,000t mA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i1(t) is,

i1(t)=i1(0+)e−tτ

Substitute 5 mA for i1(0+) and 50 μs for τ.

i1(t)=5e−t50 μs mA=5e−20,000t mA

Conclusion:

Therefore, the expression i1(t) for t≥0 is 5e−20,000t mA.

d.

To determine

Find the expression i2(t) for t≥0.

d.

Expert Solution

Answer to Problem 1P

The expression i2(t) for t≥0 is −5e−20,000t mA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i2(t) is,

i2(t)=i2(0+)e−tτ

Substitute –5 mA for i1(0+) and 50 μs for τ.

i2(t)=−5e−t50 μs mA=−5e−20,000t mA

Conclusion:

Therefore, the expression i2(t) for t≥0 is −5e−20,000t mA.

e.

To determine

Explain the reason for why i2(0−)≠i2(0+).

e.

Expert Solution

Explanation of Solution

Calculation:

The current in the resistor changes instantaneously. The switching operation makes the current i2(0−) is equal to 15 mA and i2(0+) is equal to –5 mA.

Conclusion:

Therefore, the reason for why i2(0−)≠i2(0+) is stated.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 7, Problem 1P

a.

To determine

Find the current values i1(0−) and i2(0−).

a.

Expert Solution

Answer to Problem 1P

The current values i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

Explanation of Solution

PSPICE Circuit:

Refer to the Figure P7.1 in the textbook.

Draw the given circuit diagram in PSPICE as shown in Figure 1.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 2

PSPICE output:

After run the PSPICE circuit a black output screen will be displayed. Right click on the mouse by keeping cursor on the output screen, click the option “Add Trace” and type the expression “-I(R3)” in trace expression box.

The current plot i1(t) is shown in Figure 3.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 3

From PSPICE output, the initial values of current are,

i1(0−)=5 mAi1(0−)=i1(0+)i1(0+)=5 mA

Similarly, type the expression “-I(R2)” in trace expression box to obtain the current i2(t).

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 4

From PSPICE output, the initial values of current are,

i2(0−)=15 mAi2(0+)=−5mA

Conclusion:

Therefore, the values of i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

b.

To determine

Find the current values i1(0+) and i2(0+).

b.

Expert Solution

Answer to Problem 1P

The current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

Explanation of Solution

Calculation:

From Figure 2 and Figure 3, the current values are,

i1(0+)=5 mAi2(0+)=−5mA

Conclusion:

Therefore, the current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

c.

To determine

Find the expression i1(t) for t≥0.

c.

Expert Solution

Answer to Problem 1P

The expression i1(t) for t≥0 is 5e−20,000t mA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i1(t) is,

i1(t)=i1(0+)e−tτ

Substitute 5 mA for i1(0+) and 50 μs for τ.

i1(t)=5e−t50 μs mA=5e−20,000t mA

Conclusion:

Therefore, the expression i1(t) for t≥0 is 5e−20,000t mA.

d.

To determine

Find the expression i2(t) for t≥0.

d.

Expert Solution

Answer to Problem 1P

The expression i2(t) for t≥0 is −5e−20,000t mA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i2(t) is,

i2(t)=i2(0+)e−tτ

Substitute –5 mA for i1(0+) and 50 μs for τ.

i2(t)=−5e−t50 μs mA=−5e−20,000t mA

Conclusion:

Therefore, the expression i2(t) for t≥0 is −5e−20,000t mA.

e.

To determine

Explain the reason for why i2(0−)≠i2(0+).

e.

Expert Solution

Explanation of Solution

Calculation:

The current in the resistor changes instantaneously. The switching operation makes the current i2(0−) is equal to 15 mA and i2(0+) is equal to –5 mA.

Conclusion:

Therefore, the reason for why i2(0−)≠i2(0+) is stated.

Answer 6

Chapter 7, Problem 1P

a.

To determine

Find the current values i1(0−) and i2(0−).

a.

Expert Solution

Answer to Problem 1P

The current values i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

Explanation of Solution

PSPICE Circuit:

Refer to the Figure P7.1 in the textbook.

Draw the given circuit diagram in PSPICE as shown in Figure 1.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 2

PSPICE output:

After run the PSPICE circuit a black output screen will be displayed. Right click on the mouse by keeping cursor on the output screen, click the option “Add Trace” and type the expression “-I(R3)” in trace expression box.

The current plot i1(t) is shown in Figure 3.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 3

From PSPICE output, the initial values of current are,

i1(0−)=5 mAi1(0−)=i1(0+)i1(0+)=5 mA

Similarly, type the expression “-I(R2)” in trace expression box to obtain the current i2(t).

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 4

From PSPICE output, the initial values of current are,

i2(0−)=15 mAi2(0+)=−5mA

Conclusion:

Therefore, the values of i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

Answer 7

Chapter 7, Problem 1P

a.

To determine

Find the current values i1(0−) and i2(0−).

Answer 8

a.

Expert Solution

Answer to Problem 1P

The current values i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

PSPICE Circuit:

Refer to the Figure P7.1 in the textbook.

Draw the given circuit diagram in PSPICE as shown in Figure 1.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 2

PSPICE output:

After run the PSPICE circuit a black output screen will be displayed. Right click on the mouse by keeping cursor on the output screen, click the option “Add Trace” and type the expression “-I(R3)” in trace expression box.

The current plot i1(t) is shown in Figure 3.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 3

From PSPICE output, the initial values of current are,

i1(0−)=5 mAi1(0−)=i1(0+)i1(0+)=5 mA

Similarly, type the expression “-I(R2)” in trace expression box to obtain the current i2(t).

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip 4

From PSPICE output, the initial values of current are,

i2(0−)=15 mAi2(0+)=−5mA

Conclusion:

Therefore, the values of i1(0−) and i2(0−) are 5 mA and 15 mA respectively.

Answer 13

b.

To determine

Find the current values i1(0+) and i2(0+).

b.

Expert Solution

Answer to Problem 1P

The current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

Explanation of Solution

Calculation:

From Figure 2 and Figure 3, the current values are,

i1(0+)=5 mAi2(0+)=−5mA

Conclusion:

Therefore, the current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

Answer 14

b.

To determine

Find the current values i1(0+) and i2(0+).

Answer 15

b.

Expert Solution

Answer to Problem 1P

The current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

From Figure 2 and Figure 3, the current values are,

i1(0+)=5 mAi2(0+)=−5mA

Conclusion:

Therefore, the current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

Answer 20

c.

To determine

Find the expression i1(t) for t≥0.

c.

Expert Solution

Answer to Problem 1P

The expression i1(t) for t≥0 is 5e−20,000t mA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i1(t) is,

i1(t)=i1(0+)e−tτ

Substitute 5 mA for i1(0+) and 50 μs for τ.

i1(t)=5e−t50 μs mA=5e−20,000t mA

Conclusion:

Therefore, the expression i1(t) for t≥0 is 5e−20,000t mA.

Answer 21

c.

To determine

Find the expression i1(t) for t≥0.

Answer 22

c.

Expert Solution

Answer to Problem 1P

The expression i1(t) for t≥0 is 5e−20,000t mA.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i1(t) is,

i1(t)=i1(0+)e−tτ

Substitute 5 mA for i1(0+) and 50 μs for τ.

i1(t)=5e−t50 μs mA=5e−20,000t mA

Conclusion:

Therefore, the expression i1(t) for t≥0 is 5e−20,000t mA.

Answer 27

d.

To determine

Find the expression i2(t) for t≥0.

d.

Expert Solution

Answer to Problem 1P

The expression i2(t) for t≥0 is −5e−20,000t mA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i2(t) is,

i2(t)=i2(0+)e−tτ

Substitute –5 mA for i1(0+) and 50 μs for τ.

i2(t)=−5e−t50 μs mA=−5e−20,000t mA

Conclusion:

Therefore, the expression i2(t) for t≥0 is −5e−20,000t mA.

Answer 28

d.

To determine

Find the expression i2(t) for t≥0.

Answer 29

d.

Expert Solution

Answer to Problem 1P

The expression i2(t) for t≥0 is −5e−20,000t mA.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2 kΩ+6 kΩ=8 kΩ

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 kΩ for Req and 400 mH for L.

τ=400 mH8 kΩ=400×10−38×103=50 μs

The expression i2(t) is,

i2(t)=i2(0+)e−tτ

Substitute –5 mA for i1(0+) and 50 μs for τ.

i2(t)=−5e−t50 μs mA=−5e−20,000t mA

Conclusion:

Therefore, the expression i2(t) for t≥0 is −5e−20,000t mA.

Answer 34

e.

To determine

Explain the reason for why i2(0−)≠i2(0+).

e.

Expert Solution

Explanation of Solution

Calculation:

The current in the resistor changes instantaneously. The switching operation makes the current i2(0−) is equal to 15 mA and i2(0+) is equal to –5 mA.

Conclusion:

Therefore, the reason for why i2(0−)≠i2(0+) is stated.

Answer 35

e.

To determine

Explain the reason for why i2(0−)≠i2(0+).

Answer 36

e.

Expert Solution

Explanation of Solution

Calculation:

The current in the resistor changes instantaneously. The switching operation makes the current i2(0−) is equal to 15 mA and i2(0+) is equal to –5 mA.

Conclusion:

Therefore, the reason for why i2(0−)≠i2(0+) is stated.

Answer 37

e.

Expert Solution

Answer 38

e.

Expert Solution

Answer 39

Expert Solution

Answer 40

Ch. 7.1 - The switch in the circuit shown has been closed...Ch. 7.1 - Prob. 2AP Ch. 7.2 - Prob. 3AP Ch. 7.2 - Prob. 4AP Ch. 7.3 - Prob. 5AP Ch. 7.3 - Prob. 6AP Ch. 7.4 - Prob. 7AP Ch. 7.4 - Prob. 8AP Ch. 7.5 - Prob. 9AP Ch. 7.5 - Prob. 10AP

Find the current values i 1 ( 0 − ) and i 2 ( 0 − ) .

Find the current values i1(0−) and i2(0−).

Answer to Problem 1P

Explanation of Solution

Find the current values i1(0+) and i2(0+).

Answer to Problem 1P

Explanation of Solution

Find the expression i1(t) for t≥0.

Answer to Problem 1P

Explanation of Solution

Find the expression i2(t) for t≥0.

Answer to Problem 1P

Explanation of Solution

Explain the reason for why i2(0−)≠i2(0+).

Explanation of Solution

Want to see more full solutions like this?

Chapter 7 Solutions