Question
Download Solution PDFIf two capacitors having capacitance of 5 μF and 10 μF respectively are connected in series across a 200 V supply, find the potential difference across each capacitor.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFKVL in Capacitor:
Consider two capacitors of capacitance C1 and C2 connected in series across supply having impedance Z1 and Z2 respectively as shown.
Applying Voltage division rule to the circuit,
The voltage across C1 is given as,
\(V_{C1}=V\times \frac{Z_1}{Z_1+Z_2}\) .... (1)
The voltage across C2 is given as,
\(V_{C1}=V\times \frac{Z_2}{Z_1+Z_2}\) .... (2)
The impedance Z1 and Z2 can be written as,
\(Z_1=\frac{1}{\omega C_1},Z_2=\frac{1}{\omega C_2}\)
Put the value of Z1 and Z2 in equation (1) and (2),
\(V_{C1}=V\times \frac{\frac{1}{\omega C_1}}{\frac{1}{\omega C_1}+\frac{1}{\omega C_2}}\)
Hence, the voltage across C2 will be,
\(V_{C_1}=V\times \frac{C_2}{C_1+C_2}\)
Similarly, Voltage across C2 will be,
\(V_{C_2}=V\times \frac{C_1}{C_1+C_2}\)
Application:
Given,
C1 = 5 μF
C2 = 10 μF
V = 200 V
From the above concept, the voltage across C1 is given as,
\(V_{C_1}=V\times \frac{C_2}{C_1+C_2}=200\times \frac{10}{10+5}=133.33 V\)
And the voltage across C2 is given as,
\(V_{C_2}=V\times \frac{C_1}{C_1+C_2}=200\times \frac{5}{10+5}=66.66 V\)
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