Question 1.SA.5: Determine the equivalent capacitance for the DC circuit show......
Determine the equivalent capacitance for the DC circuit shown in the circuit diagram below if C_1 = 5 μF and C_2 = 10 μF.
Step-by-Step
Application of Eq. 1.21 to the two-capacitor series circuit shown in the given circuit diagram yields:
C_{EQ}=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+…+\frac{1}{C_n}} (1.21)
C_{EQ}=\frac{C_1C_2}{C_1+C_2}\\ C_{EQ}=\frac{(5×10^{-6})(10×10^{-6})}{(5×10^{-6})+(10×10^{-6})}=3.33×10^{-6}=3.33 μF
Related Answered Questions
Question: 1.SA.10
Verified Answer:
Apply series RL current equation, Eq. 1.31.
[latex...
Question: 1.SA.7
Verified Answer:
The capacitors in this circuit that lend themselve...
Question: 1.SA.8
Verified Answer:
If the DC source is replaced by an AC source, the ...
Question: 1.SA.9
Verified Answer:
In most series RL cases, the current value at a ce...
Question: 1.SA.13
Verified Answer:
If the DC source is replaced by an AC source, the ...
Question: 1.SA.11
Verified Answer:
Apply Eq. 1.36 to compute L_{EQ} fo...
Question: 1.SA.12
Verified Answer:
Focus on the parallel combination of L_2, L...
Question: 1.SA.4
Verified Answer:
This particular case represents a capacitor chargi...
Question: 1.SA.6
Verified Answer:
Apply Eq. 1.22 to “n” series capacitor circuit sho...
Question: 1.SA.3
Verified Answer:
No current will flow through the two 4 Ω resistors...