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Solved Problems-16

Problems-16
A 50μF capacitor and 20000Ω resistor are connected in series across a 100V battery at t = 0. At t = 0.5s, the battery voltage is suddenly increased to 150V. Find the charge on capacitor at t = 0.75s.
Solution
When the circuit is connected to 100V supply, the equation of voltage across the capacitor is,
 
Description: Description: 8831.png
 
At t = 0.5s, the voltage across the capacitor is, Description: Description: 8840.png
 
∴ At t = 0.5s, charge on the capacitor is, q = CvC = 50 × 10–6 × 39.347 = 1967.35 × 10–6 C
This charge is the initial charge q0 when the battery voltage is suddenly increased to 150V.
When the circuit is connected to 150V, the KVL equation becomes,
Description: Description: 8861.png
 
Taking Laplace transform,
 
Description: Description: 8867.png
⇒ Description: Description: 8873.png
 
Taking inverse Laplace transform,
 
Description: Description: 8879.png
 
Therefore, the voltage across the capacitor,
 
Description: Description: 8885.png
 
Substituting the values, the voltage across the capacitor at t = 0.75s i.e., 0.25 second after changing the battery voltage,
 
Description: Description: 8891.png
 
∴ charge on the capacitor is, Description: Description: 8897.png
 




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