Consider a circuit that has two capacitors in it.?The first has a capacitance of

Question

Consider a circuit that has two capacitors in it.?The first has a capacitance of 4x10^(-9) F and the second has a capacitance of 5x10^(-9) F . The first capacitor has a dielectric placed between the plates with dielectric constant of 1.5 The second capacitor has a dielectric placed between the plates with dielectric constant of 2. What would the equivalent capacitance be if the capacitors were placed: a)in series B) in parallel c) If while connected in parallel and hooked up to a battery, the first capacitor holds a charge of 10C, what is the voltage supplied by the battery?

Explanation / Answer

Given;-

C1=4x10-9F

C2=5x10-9F

K1=1.5

K2=2

a] The equivalent capacitance be if the capacitors were placed: in series

Cequivalent = k1c1K2c2/ [K1c1+K2c2] =1.5*4*2*5*10-18/[1.5*4 + 2*5]10-9 = 3.75*10 -9F

b] The equivalent capacitance be if the capacitors were placed: in parallel

Cequivalent = k1c1+K2c2 = [1.5*4 + 2*5]10-9 = 16*10-9 F

c]

q=10C

Voltage in parallel mode remains same for the two capacitors

q= cv

v= q/c=10 C/16*10-9 F=6.25*108 V

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