Problem design:

For the circuit in Figure 6.57, determine the voltage across each capacitor and the energy stored in each capacitor.

Formula used:

Write the expression to calculate the energy stored in the capacitor.

w=12CV2        (1)

Here,

C is the value of the capacitor, and

V is the voltage across the capacitor.

Calculation:

Refer to Figure 6.57 in the textbook. The Figure 6.57 is redrawn as Figure 1 by assuming the voltage and capacitor values.

Refer to Figure 1, the capacitors C3 and C4 are connected in series form.

Write the expression to calculate the equivalent capacitance 1 for the series connected capacitors C3 and C4.

Ceq1=C3C4C3+C4        (2)

Here,

C3 is the value of the capacitor 3, and

C4 is the value of the capacitor 4.

Substitute 6 μF for C3 and 3 μF for C4 in equation (2) to find Ceq1.

Ceq1=(6 μF)(3 μF)6 μF+3 μF=18 (μF)29 μF=2 μF

The reduced circuit of the Figure 1 is drawn as Figure 2.

Refer to Figure 2, the capacitors C2 and Ceq1 are connected in parallel form.

Write the expression to calculate the equivalent capacitance 2 for the parallel connected capacitors C2 and Ceq1.

Ceq2=C2+Ceq1        (3)

Here,

C2 is the value of the capacitor 2, and

Ceq1 is the value of the equivalent capacitance 1.

Substitute 2 μF for C2 and 2 μF for Ceq1 in equation (3) to find Ceq2.

Ceq2=2 μF+2 μF=4 μF

The reduced circuit of the Figure 2 is drawn as Figure 3.

Write the expression to calculate the total applied voltage.

V=V1+V2        (4)

Here,

V1 is the voltage across the capacitor 1, and

V2 is the voltage across the equivalent capacitor 2.

From the Figure 3, it is clear that the voltage across the capacitors with same capacitance value is equal. Therefore,

V1=V2        (5)

Substitute 120 V for V and V1 for V2 in equation (4).

120 V=V1+V1=2V1

Simplify the above equation to find V1.

V1=120 V2=60 V

Therefore, from equation (5),

V1=V2=60 V

Write the expression to calculate the charge across the capacitor 1.

q1=C1V1        (6)

Here,

C1 is the value of the capacitor 1.

Substitute 4 μF for C1 and 60 V for V1 in equation (6) to find q1.

q1=(4 μF)(60 V)=(4×10−6)(60) F⋅V {∵1 μ=10−6}=0.24×10−3 (CV)⋅V {∵1F=1 C1V}=0.24 mC {∵1 m=10−3}

Write the expression to calculate the charge across the capacitor 2.

q2=C2V2        (7)

Substitute 2 μF for C2 and 60 V for V2 in equation (7) to find q2.

q2=(2 μF)(60 V)=(2×10−6)(60) F⋅V {∵1 μ=10−6}=0.12×10−3 (CV)⋅V {∵1F=1 C1V}=0.12 mC {∵1 m=10−3}

Refer to Figure 2, the capacitors C2 and Ceq1 are connected in parallel form. For parallel connection, the voltage is same. Here, the capacitors C2 and Ceq1 have same capacitance value and voltage, therefore the charge of the capacitors should be same that is charge q2.

The combination of the series connected capacitors C3 and C4 is Ceq1. For series connection, the charge is same. Therefore, the charge for the capacitors C3 and C4 is same as Ceq1 that is q2.

q2=q3=q4

Write the expression to calculate the voltage across the capacitor 3.

V3=q2C3        (8)

Substitute 0.12 mC for q2 and 6 μF for C3 in equation (8) to find V3.

V3=0.12 mC6 μF=0.12× 10−36×10−6 CF {∵1 m=10−3, 1 μ=10−6}=20 (V⋅FF) {∵1 C=1 V⋅1 F}=20 V

Write the expression to calculate the voltage across the capacitor 4.

V4=q2C4        (9)

Substitute 0.12 mC for q2 and 3 μF for C4 in equation (9) to find V4.

V4=0.12 mC3 μF=0.12× 10−33×10−6 CF {∵1 m=10−3, 1 μ=10−6}=40 (V⋅FF) {∵1 C=1 V⋅1 F}=40 V {∵1 V=1C1F}

Re-write the equation (1) to calculate the energy stored in capacitor 1.

w1=12C1V12

Substitute 4 μF for C1 and 60 V for V1 in above equation to find w1.

w1=12(4 μF)(60 V)2=12(4×10−6)(60)2 F⋅V2 {∵1 μ=10−6}=7.2×10−3 (JV2)⋅V2 {∵1 F=1 J1 V2}=7.2 mJ {∵1 m=10−3}

Re-write the equation (1) to calculate the energy stored in capacitor 2.

w2=12C2V22

Substitute 2 μF for C2 and 60 V for V2 in above equation to find w2.

w2=12(2 μF)(60 V)2=12(2×10−6)(60)2 F⋅V2 {∵1 μ=10−6}=3.6×10−3 (JV2)⋅V2 {∵1 F=1 J1 V2}=3.6 mJ {∵1 m=10−3}

Re-write the equation (1) to calculate the energy stored in capacitor 3.

w3=12C3V32

Substitute 6 μF for C3 and 20 V for V3 in above equation to find w3.

w3=12(6 μF)(20 V)2=12(6×10−6)(20)2 F⋅V2 {∵1 μ=10−6}=1.2×10−3 (JV2)⋅V2 {∵1 F=1 J1 V2}=1.2 mJ {∵1 m=10−3}

Re-write the equation (1) to calculate the energy stored in capacitor 4.

w4=12C4V42

Substitute 3 μF for C4 and 40 V for V4 in above equation to find w4.

w4=12(3 μF)(40 V)2=12(3×10−6)(40)2 F⋅V2 {∵1 μ=10−6}=2.4×10−3 (JV2)⋅V2 {∵1 F=1 J1 V2}=2.4 mJ {∵1 m=10−3}

Therefore, the value of the voltage across the capacitor 1 V1 is 60 V, the capacitor 2 V2 is 60 V, the capacitor 3 V3 is 20 V and the capacitor 4 V4 is 40 V and the energy stored in capacitor 1 w1 is 7.2 mJ, the capacitor 2 w2 is 3.6 mJ, the capacitor 3 w3 is 1.2 mJ and the capacitor 4 w4 is 2.4 mJ.

Conclusion:

Thus, the problem to make better understand how capacitors work together when connected in series and in parallel using Figure 6.57 is designed.