4) Consider the circuit shown below. find the power delivered to each resistor.

Question

4) Consider the circuit shown below. find the power delivered to each resistor.

5) Consider the circuit shown at right let e=40v, R1=8 omega, R2....

a) find the current through each resistor immediately after the switch is closed

b) find the final charge on the capacitor

6) A 12-volt battery charges the four capacitors shown below. Let C1= 1 mu F, ....

a) what is the equivalent capacitance of the group C1 and C2 if swithc S is open (as shown)?

b) what is the charge on each of the four capacitors if switch S is open?

c) what is the charge on each of the four capacitors if switch S is closed?

Explanation / Answer

4ans) power= i^2*R

R-resistance

i-current

equivalent Resitance =2+4+(3/4)=6.75 ohms

current= voltage / resitance =20/6.75=2.963 amperes

therefore power in 2 ohm resistor=2.963^2*2=17.558watts

power in 4 ohm resistor=2.963^2*4=35.116watts

in parralel resistance current gets divide in the inverse ratio of resistance

therefore current in 1ohm =2.693*(3/4)

therefore current in 3ohm =2.693*(1/4)

power in 3 ohm resistor = (2.693*(1/4))^2*3=1.36watts   

power in 1 ohm =(2.693*(3/4))^2*1=4.08watts

5ans) immediately after switch closed capacitor acts as closed circuit witth zero resistance

paraller resistance equivalent 1/R=1/6+1/4=1/2.4 therefore R=2.4 ohms

equivalent resistance =8+2.4=10.4ohms

current=40/10.4=3.846amps

current in 8 ohms =3.846amps

currennt in 6 ohms=3.846*(4/10)=1.5384amps

current in 4 ohms =3.846*(6/10)=2.3076amps

after capacitor charged completely it acts as open circuit with zero current

voltage of capacitor = voltage across R2=40*(6 / 8+6)=40*6/14=17.14v

charge = capacitance *voltage=4*10^-6 * 17.14=6.857*10^-7 coloumbs

6ans) C1 C2 equivalent C

1/C=1+1/2=3/2 immplies C=2/3 micro farads

the voltage across the capacitors will be distributed according to the following formulas.
. Vc1 = V C2/(C1 + C2)
. Vc2 = V C1/(C1 + C2)
Where
. Vc1 is the voltage across C1
. Vc2 is the voltage across C2
. V is the voltage applied across the series combination=12V

therfore charge on C1, C2 capacitors Q1,Q2 =(V*C1*C2)/(C1+C2)=12* 2/3 *10^-6=8*10^-6coloumbs

therfore charge on C3, C4 capacitors Q3,Q4 =(V*C3*C4)/(C3+C4)=12* 12/7 *10^-6=20.57*10^-6coloumbs

switch closed

C1 C3 equivalent capacitance Cx=C1+C3=4uF

C2 C4 equivalent capacitance Cy=C2+C4=6uF

Vcx = V Cy/(Cx + Cy) =12*0.6=7.2v
. Vcy = V Cx/(Cx + Cy)=12*.4=4.8v
Where
. Vcx is the voltage across Cx
. Vcy is the voltage across Cy

Q1=C1*Vx=7.2*10^-6 couloumbs

Q3=C3*Vx=3*7.2*10^-6=21.6*10^-6 couloumbs

Q2=C2*Vy=2*4.8*10^-6 =9.6*10^-6couloumbs

Q4=C4*Vy=4*4.8*10^-6 =19.2*10^-6couloumbs

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