For the circuit (Figure 1), take E = 400 V, R1=4.0kΩ, R2=6.0kΩ, and assume the capacitor is initially uncharged. 1. What are the currents in both resistors just after the switch is closed? (I1,I2)
2. What is the voltage across the capacitor just after the switch is closed?

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energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j