Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 63)
63.
What is the voltage across the capacitor?
Discussion:
9 comments Page 1 of 1.
Ristoria said:
5 years ago
The problem clearly states that the voltage drop on R is 7.072 V via the Voltmeter reading which means current flowing through the resistor would be 7.072mA and since it is a SERIES circuit current is the same for the resistor and capacitor, therefore, the VC = (7.072mA)(15.92ohms) = 112.6mV, 111mV is the closest.
(1)
Saurav Bhagabati said:
1 decade ago
I need the explanation to this problem.
RESHMA said:
1 decade ago
i=7.072/1000=.007072A
Xc=1/2PiFC=1/2*3.14*10*10^3*10^-6=15.9 ohm
Voltage across capacitor =I*Xc=.007072*15.9=.112V
So Approximately =111 mV.
Xc=1/2PiFC=1/2*3.14*10*10^3*10^-6=15.9 ohm
Voltage across capacitor =I*Xc=.007072*15.9=.112V
So Approximately =111 mV.
Girish said:
1 decade ago
How about?
vsuppy2=vcapacitor2+vresistor2
So v capacitor is also equal to around 7 volt.
vsuppy2=vcapacitor2+vresistor2
So v capacitor is also equal to around 7 volt.
MUKESH kumar said:
1 decade ago
vrms=10/1.414 VR1+VC1=VRMS
Leo said:
1 decade ago
Considering the given R=1K, C=1uF, V1=10V/10Khz, and all is RMS voltage, the current is:
i=V1/Z which is 10/sqrt(1000^2+15,9^2) = 9,99 mA and not 7,072 mA
Therefor Uc=9,99E-3*15,91=0,159 V or 159 mV
i=V1/Z which is 10/sqrt(1000^2+15,9^2) = 9,99 mA and not 7,072 mA
Therefor Uc=9,99E-3*15,91=0,159 V or 159 mV
Kram said:
1 decade ago
I agree with @Leo, but what about the formula Vt = sqrt((Vr*Vr)+(Vc*Vc))??
Ninzkie said:
1 decade ago
Yes what about this relation?
10(squared) = Vr(squared) + Vc(squared)?
10(squared) = Vr(squared) + Vc(squared)?
Colin Mitchell said:
2 years ago
How can you have a positive and negative on an AC voltage?
Is the voltmeter reading AC?
Please, anyone, explain to me.
Is the voltmeter reading AC?
Please, anyone, explain to me.
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