Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 31)
31.
If C1, a 4.7 
F capacitor, and C2, a 3.3 F capacitor are in series with 18 Vdc applied, what is the voltage across C1?
F capacitor, and C2, a 3.3 F capacitor are in series with 18 Vdc applied, what is the voltage across C1?Discussion:
17 comments Page 2 of 2.
Mahadev said:
1 decade ago
All 18 Vdc should be across c1 because for dc capacitor will be open circuit.
Yamini said:
1 decade ago
Capacitor acts as open circuit for dc current because frequency f=0 in dc current.
Archan mani said:
1 decade ago
Only ideal frequency zero in dc but practicaly some frequency is available.
Samay said:
1 decade ago
@vishwanath : i also wondering the same
Vishwanath said:
1 decade ago
How can a dc current flow through a capacitor ?
M.V.KRISHNA said:
1 decade ago
Given
C1=4.7uF, C2=3.3uF, V=18dc;
two capacitors are in series so
Cs=c1*c2/(c1+c2);
Cs=4.7uF * 3.3uF/(4.7uF + 3.3uF);
Cs=1.93uF;
voltage across V1= VCs/C1;
=> 18*1.93/4.7
=7.4V
C1=4.7uF, C2=3.3uF, V=18dc;
two capacitors are in series so
Cs=c1*c2/(c1+c2);
Cs=4.7uF * 3.3uF/(4.7uF + 3.3uF);
Cs=1.93uF;
voltage across V1= VCs/C1;
=> 18*1.93/4.7
=7.4V
Pankaj garg said:
1 decade ago
Total Capacitance Ct = c1*c2/c1+c2 =
C=g/V
We find out the charge
q=Ct*V
THEN WE Find out the voltage across C1
V1= q/c1
C=g/V
We find out the charge
q=Ct*V
THEN WE Find out the voltage across C1
V1= q/c1
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