Find the energy stored in the element at time t = 0 for the circuit shown in Figure 7.83.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 7, Problem 69E

(a)

To determine

Find the energy stored in the element at time t=0 for the circuit shown in Figure 7.83.

(a)

Expert Solution

Answer to Problem 69E

The energy stored in the element at time t=0 for the circuit shown in Figure 7.83 is 405 μJ .

Explanation of Solution

Formula used:

Write a general expression to calculate the energy stored in an capacitor.

w=12Cv2(t) (1)

Here,

C is the value of capacitance, and

v(t) is the voltage across the capacitor.

Given data:

Refer to Figure 7.83 in the textbook.

The value of initial voltage across the capacitor vC(0)=v(0) is 9 V.

Calculation:

Substitute 0 for t in equation (1) to find w at time t=0.

w=12C(v(0))2

Substitute 10 μF for C and 9 V for v(0) in equation (1) to find w.

w=12(10 μF)(9 V)2=12(10×10−6 F)(81 V2) {∵1 μ=10−6}=405×10−6 F⋅V=405×10−6 (A⋅sV⋅V2) {∵1 F=1 A⋅1 s1 V}

Simplify the above equation to find w.

w=405×10−6 (A⋅s⋅V)=405×10−6 (Cs⋅s⋅V) {∵1 A=1 C1 s}=405×10−6 C⋅V=405 μJ {∵1 J=1 C⋅1 V 1 μ=10−6}

Conclusion:

Thus, the energy stored in the element at time t=0 for the circuit shown in Figure 7.83 is 405 μJ .

(b)

To determine

Explain whether the energy stored in the capacitor remains same for time t>0.

(b)

Expert Solution

Answer to Problem 69E

No, the energy is not same in the capacitor for time t>0.

Explanation of Solution

Refer to Figure 7.83, it shows an RC circuit. When an DC voltage is applied to the capacitor the charges up with an increasing voltage. Since the energy of capacitor is directly proportional to the voltage, for an increasing voltage energy also increases.

Due to the presence of resistor in a circuit, there is a voltage drop across the resistor which reduces the voltage across the capacitor. Therefore, the resistor slowly dissipate the energy stored in a capacitor for time t>0.

Conclusion:

Thus, the energy is not same in the capacitor for time t>0.

(c)

To determine

Determine the value of v(t) at time t=460 ms, t=920 ms, and t=2.3 s using transient simulation.

(c)

Expert Solution

Answer to Problem 69E

The value of v(t) at time t=460 ms is, t=920 ms is, and t=2.3 s is

Explanation of Solution

Calculation:

Create the new schematic in LTspice with series connected resistor and inductor of given circuit as shown in Figure 1.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 7, Problem 69E , additional homework tip 1

Using SPICE Directive mention the command .ic V(Cap_voltage)=9 as shown in Figure 2.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 7, Problem 69E , additional homework tip 2

Enter the stop time as 2.5s, time to start saving data as 0, and maximum Timestep as 10ms in Edit simulation Cmd as shown in Figure 3. Use label net option and mention Cap_voltage.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 7, Problem 69E , additional homework tip 3

After adding the Spice directives the circuit shows as in Figure 4.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 7, Problem 69E , additional homework tip 4

Now run the simulation and place the probe at the node of capacitor, the plot of the voltage across the capacitor with respect to time is shown as shown in Figure 5.

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf, Chapter 7, Problem 69E , additional homework tip 5

By placing the cursor on the graph, we obtain the current values for different time as shown in below.

For time t=0 s v(t)=9 V

For time t=460 ms v(t)=3.31 V

For time t=920 ms,v(t)=1.22 V

For time t=2.3 s, v(t)=60.4 mV

Conclusion:

Thus, the value of v(t) at time t=460 ms is 3.31 V, t=920 ms is 1.22 V, and t=2.3 s is 60.4 mV.

(d)

To determine

Find the fraction of initial energy remains in the capacitor at time t=460 ms and t=2.3 s.

(d)

Expert Solution

Answer to Problem 69E

The fraction of initial energy remains in the capacitor at time t=460 ms is 13.6% and t=2.3 s is 0.0045%.

Explanation of Solution

Calculation:

Refer to part (b), the value of voltage at time t=460 ms is v(460 ms)=3.31 V and the value of voltage at time t=2.3 s is v(2.3 s)=60.4 mV

Substitute 460 ms for t in equation (1) to find w at time t=460 ms.

w(460 ms)=12C(v(460 ms))2

Substitute 3.31 V for v(460 ms), and 10 μF for C in above equation to find w(460 ms).

w(460 ms)=12(10 μF)(3.31 V)2=12(10×10−6 F)(11 V2) {∵1 μ=10−6}=55×10−6 F⋅V2 {∵1 μ=10−6}=55×10−6 (A⋅sV⋅V2) {∵1 F=1 A⋅1 s1 V}

Simplify the above equation to find w(460 ms).

w(460 ms)=55×10−6 (A⋅s⋅V)=55×10−6 (Cs⋅s⋅V) {∵1 A=1 C1 s}=55×10−6 C⋅V=55 μJ {∵1 J=1 C⋅1 V 1 μ=10−6}

Substitute 2.3 s for t in equation (1) to find w at time t=2.3 s.

w(2.3 s)=12C(v(2.3 s))2

Substitute 60.4 mV for v(2.3 s), and 10 μF for C in above equation to find w(2.3 s).

w(2.3 s)=12(10 μF)(60.4 mV)2=12(10×10−6 F)(60.4×10−3 V)2 {∵1 μ=10−6 1 m=10−3}=12(10×10−6 F)(3.648×10−3 V2)=18.2×10−9 F⋅V2

Simplify the above equation to find w(2.3 s).

w(2.3 s)=18.2×10−9 (A⋅sV⋅V2) {∵1 F=1 A⋅1 s1 V}=18.2×10−9 (A⋅s⋅V)=18.2×10−9 (Cs⋅s⋅V) {∵1 A=1 C1 s}=18.2×10−9 C⋅V

w(2.3 s)=18.2 nJ {∵1 J=1 C⋅1 V 1 n=10−9}

The fraction of initial energy remains stored in the capacitor at time t=460 ms is calculated as follows:

Fraction of energy at t=460 ms=w(460 ms)w(0)×100%

Substitute 405 μJ for w(0), and 55 μJ for w(460 ms) in above equation to find Fraction of energy at t=460 ms.

Fraction of energy at (t=460 ms)=55 μJ405 μJ×100%=13.6%

The fraction of initial energy remains stored in the capacitor at time t=2.3 s is calculated as follows:

Fraction of energy at t=2.3 s=w(2.3 s)w(0)×100%

Substitute 405 μJ for w(0), and 18.2 nJ for w(2.3 s) in above equation to find Fraction of energy at t=2.3 s.

Fraction of energy at( t=2.3 s)=18.2 nJ405 μJ×100%=18.2×10−9 J405×10−6 J×100%=0.0045%

Conclusion:

Thus, the fraction of initial energy remains in the capacitor at time t=460 ms is 13.6% and t=2.3 s is 0.0045%.

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

The switch in the circuit shown in Fig. 7.21 has been in position a for a longtime. At t=0, the switch moves from position a to position b. The switch is amake-before-break type; that is, the connection at position b is establishedbefore the connection at position a is broken, so the inductor current iscontinuous.5. Plot both i(t) and v(t) versus t.

In response to a change introduced by a switch at t = 0, the current flowing through a 100 μF capacitor, defined in accordance with the passive sign convention, was observed to be i(t) = −0.4e−0.5t mA (for t > 0). If the final energy stored in the capacitor (at t = ∞) is 0.2 mJ, determine υ(t) for t ≥ 0.

Find Leq between the terminals a,b for the circuits shown below. Assuming the initial energy stored in the inductors is zero.

Answer 2