![]() ![]() Similarly, if the net capacities of C two and C three connected in series is C dash, then the net voltage across see dash is also equal to 12 volts. And that is very evident from the it is very evident from The figure here that C1, the voltage across the one is this is the voltage across these two again, 12 volt. The net voltage across each of the capacitor that we know that the voltage across capacitor C1 is equal to 12 votes. So the third part of the question we're asked to find out what is the next voltage across each of the capacitor. Now we will move on to the third part of the question. So this is the total amount of energy that is stored in the circuit. This isn't jews and this we get to be equal to Okay. ![]() ![]() Multiplied by 10 to the power minus silks minus multiplied by 144. So if you make this calculation, we get the total energy stored In the capacity is equal to one x 2. So we'll find out the energy stored in the circuit in the circuit is you is equal to have seen it V. So in the second part of the question we are asked to find out what is the net energy stored in the circuit. Now we'll move on to the second part of the question. So this is the value of net capacitance of the circuit. That is equal to 1.506 plus The value of C1 we have is equal to 4.86 4.86 micro Ferrari and this is equal to 6.366 micro for that. So we have the net capacities of the circuit is equal to c one plus c dash. So we have cedars, ad C one r parallel to each other are parallel to each other. And this cedars capacitor is then connected in parallel to the C one. ![]() On making this calculation we get the value of c dash to be equal to 1.506 Micro Farhad. Or we have C dash is equal to mhm C two multiplied by C3, divided by C two plus C three Supporting all the valleys we get C2 multiplied by C3 is 2.39, multiplied by 4.07 divided by 2.39 plus 4.07 micro farhad. So we have if Cicadas is the net capacity of C2 and C3 connected in series then see dash is equal to so one by CDA should be equal to one by C two plus one by C three. So first question is that we need to find out the net capacities of the circuit so we can see that capacitor, C two and C three are connected in series are connected in series. So in this question We have the following circuit consisting of various capacities connected across a 12 volt battery. ![]()
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