Question

plz ans. following questions.
1) Define the physical quantity whose unit of J/c.
2) How many Electron volt make one joule ?
3) what is one Picofarad?
4) Find an expression for equivalent capacity when a few conductors are connected in parallel.
5) Drive am the expression for the net capacitance of three capacitor in series combination.
(I Will Mark As a brain list)
please Ans. fastly plzzzzzzzzzz ♥♥♥ ♥♥♥

Answer 1

1.the physical quantity which has its unit joule coulomb–1. ... Poetential is defined as work done per unitcharge. This implies, Joule per Coulomb is the SI unit of electric potential.

2.Electron volt, unit of energy commonly used in atomic and nuclear physics, equal to the energy gained by an electron (a charged particle carrying unit electronic charge) when the electrical potential at the electronincreases by one volt. The electron volt equals 1.602 × 10−12 erg, or 1.602 × 10−19 joule.

3.1 μF (microfarad, one millionth (10−6) of a farad) = 0.000 001 F = 1000 nF = 1000000 pF. 1 nF (nanofarad, one billionth (10−9) of a farad) = 0.001 μF = 1000 pF. 1 pF (picofarad, one trillionth (10−12) of a farad)

4.Equivalent parallel conductances. We can imagine a new conductance equivalent to the sum of the parallel conductances. It is equivalent in the sense that the same voltage appears.

5.When capacitors are connected in series, the total capacitance is less than any one of the series capacitors' individual capacitances. ... If two or more capacitors are connected in parallel, the overall effect is that of a single equivalent capacitor having the sum total of the plate areas of the individual capacitors.

5.Taking the three capacitor values from the above example, we can calculate the total capacitance, CT for the three capacitors in series as:


One important point to remember about capacitors that are connected together in a series configuration, is that the total circuit capacitance ( CT ) of any number of capacitors connected together in series will always be LESS than the value of the smallest capacitor in the series and in our example above CT = 0.055μF with the value of the smallest capacitor in the series chain is only 0.1μF.
This reciprocal method of calculation can be used for calculating any number of individual capacitors connected together in a single series network. If however, there are only two capacitors in series, then a much simpler and quicker formula can be used and is given as:



If the two series connected capacitors are equal and of the same value, that is: C1 = C2, we can simplify the above equation further as follows to find the total capacitance of the series combination.


Then we can see that if and only if the two series connected capacitors are the same and equal, then the total capacitance, CT will be exactly equal to one half of the capacitance value, that is: C/2.
With series connected resistors, the sum of all the voltage drops across the series circuit will be equal to the applied voltage VS( Kirchhoff’s Voltage Law ) and this is also true about capacitors in series.
With series connected capacitors, the capacitive reactance of the capacitor acts as an impedance due to the frequency of the supply. This capacitive reactance produces a voltage drop across each capacitor, therefore the series connected capacitors act as a capacitive voltage divider network.
The result is that the voltage divider formula applied to resistors can also be used to find the individual voltages for two capacitors in series. Then:



Where: CX is the capacitance of the capacitor in question, VS is the supply voltage across the series chain and VCX is the voltage drop across the target capacitor.

Answer 2

ans 1= the physical quantity is potential(v)

v=work/charge

ans 3=picofarad is of order 10^-12 farad

ans 4= so let us consider  2 capacitors c1 and c2 connected in parallel

so,

we know that q=cv

and in parallel potential remains same charge divides

charge on first capacitor c1

q1=c1v

similarly

q2=c2v

let net charge be q

q(net)=c(net)v

q=q1+q2

so final expression

c(net)v=c1v+c2v

v cancels out from lhs and rhs

so c(net)=c1+c2

ans 5= again let us consider two capacitors c1 and c2 connected in series

so charge remains same and voltage divides

so let net voltage be v

v(net)=q/c(net)

so

for first capacitor and second capacitor

v1=q/c1

v2=q/c2

v(net)=v1+v2

q/c(net)=q/c1+q/c2

charge cancels out so

1/c(net)=1/c1+1/c2

so final exp