What is the expression for the resultant value of three capacitance where C1 connected in parallel

Computers and Technology

1 answer:

harina [27]3 years ago

7 0

Explanation:

Consider three capacitors of capacitance C1, C2, and C3 connected in series with a battery of voltage V as shown in figure (a). As soon as the battery is connected to the capacitors in series, the electrons of charge -Q are transferred from the negative terminal to the right plate of C3 which pushes the electrons of the same amount -Q from a left plate of C3 to the right plate of C2 due to electrostatic induction. Similarly, the left plate of C2 pushes the charges of Q to the right plate of which induces the positive charge +Q on the left plate of C1 At the same time, electrons of charge -Q are transferred from the left plate of C1 to the positive terminal of the battery.

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Write a statement that defines a double variable named result, initialized with the value 6.759.

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Answer:

"double result=6.759;" is the correct answer for the above question.

Explanation:

In c-programming language, the double is a data type which stores the decimal value up to 6 decimal point.
This data type takes 8 bytes space in memory when it is used in the c-programming language.
When the user wants to declare the double data type, then he should need to declare by the help of the following syntax-- double variable_name_or _identifier_name;
When the user wants to declare the double data type and initialize the value on its then he can do this with the help of the following syntax--double variable_name_or _identifier_name= value_which_needs_to_store;
The above question asked the one-line statement which declares the result variable of double data type and initializes the "6.759" value on its then he can do this by the help of "double result=6.759;" statement which is described above. Hence the answer is "double result=6.759;".