Please help with capacitance problem. Simplify as much as possible
please!Homework Answers
90 v is applied along two parallel path. so voltage along each path is 90 v. so voltage across 6uF capacitor is 90*(3/(3+6))=30 v
Qeq=299.7 uC
Let q1 in upper line q1/1.33 = 299.7-q1 / 2 q1=119.7uC
therfore charge in 6uF = 180uC
Voltage (6uF) = 180/6 =30 Volts
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Please help with capacitance problem. Simplify as much as possible
please!
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Question 5
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Question 5
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