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

Question

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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.

EBK ELECTRIC CIRCUITS, Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

EBK ELECTRIC CIRCUITS, 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.

EBK ELECTRIC CIRCUITS, 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).

EBK ELECTRIC CIRCUITS, 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.

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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.

EBK ELECTRIC CIRCUITS, Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

EBK ELECTRIC CIRCUITS, 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.

EBK ELECTRIC CIRCUITS, 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).

EBK ELECTRIC CIRCUITS, 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.

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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.

EBK ELECTRIC CIRCUITS, Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

EBK ELECTRIC CIRCUITS, 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.

EBK ELECTRIC CIRCUITS, 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).

EBK ELECTRIC CIRCUITS, 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.

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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.

EBK ELECTRIC CIRCUITS, Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

EBK ELECTRIC CIRCUITS, 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.

EBK ELECTRIC CIRCUITS, 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).

EBK ELECTRIC CIRCUITS, 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.

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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.

EBK ELECTRIC CIRCUITS, Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

EBK ELECTRIC CIRCUITS, 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.

EBK ELECTRIC CIRCUITS, 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).

EBK ELECTRIC CIRCUITS, 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.

EBK ELECTRIC CIRCUITS, Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

EBK ELECTRIC CIRCUITS, 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.

EBK ELECTRIC CIRCUITS, 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).

EBK ELECTRIC CIRCUITS, 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.

EBK ELECTRIC CIRCUITS, Chapter 7, Problem 1P , additional homework tip 1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

EBK ELECTRIC CIRCUITS, 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.

EBK ELECTRIC CIRCUITS, 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).

EBK ELECTRIC CIRCUITS, 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

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