Calculate the potential difference and the energy stored in the capacitor c2 in the circuit
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Three capacitors c1=20×10-6f, c2=30×10-6f, c3=15×10-6f are respectively connected in series. Therefore according to series connection of capacitors,
1c=120+130+11510-6=3+2+46010-6=960×10-6 or
c=609×10-6=203×10-6f.
Here c is net effective capacitance of the circuit. Suppose total charge Qflows through the network of capacitors connected in series.
So, Q=CV Here V=90V, therefore, Q=203×10-6×90=6×10-4C.Now potential difference across c2 is,V2=Qc2 or V2=6×10-430×10-6=20V.The energy stored by the capacitor c2=12c2v22=12×30×10-6×202=15×400×10-6=6×10-3J.
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1c=120+130+11510-6=3+2+46010-6=960×10-6 or
c=609×10-6=203×10-6f.
Here c is net effective capacitance of the circuit. Suppose total charge Qflows through the network of capacitors connected in series.
So, Q=CV Here V=90V, therefore, Q=203×10-6×90=6×10-4C.Now potential difference across c2 is,V2=Qc2 or V2=6×10-430×10-6=20V.The energy stored by the capacitor c2=12c2v22=12×30×10-6×202=15×400×10-6=6×10-3J.
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