Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 1)
1.
In a 20 Vac series RC circuit, if 20 V is measured across the resistor and 40 V is measured across the capacitor, the applied voltage is:
Discussion:
37 comments Page 1 of 4.
Merine said:
1 decade ago
How the problem can be solved?
Debadatta said:
1 decade ago
What is the solution of this question?
MUNIRAJ said:
1 decade ago
How the problem can be solved?
Jayendra said:
1 decade ago
Here, rc series circuit therefore as formula vt= squareroot of vr*vr+vc*vc=400.
1600=44. 72=45v.
Thats the answer.
1600=44. 72=45v.
Thats the answer.
Vadivelan said:
1 decade ago
Ya above method is correct.
Manoj said:
1 decade ago
I agree with above statement.
Chandrasekhar said:
1 decade ago
Let us for suppose that an i current is flowed in to the circuit such that "the voltage developed across the series pair of resistance and capacitor is == I(R+jXc)
The voltage = IR + j(IXc) = voltage acc resistance + j(voltage acrosscapacitor)
= 20+j40
= sqrt(20*20 + 40 *40)
= 44.72
The voltage = IR + j(IXc) = voltage acc resistance + j(voltage acrosscapacitor)
= 20+j40
= sqrt(20*20 + 40 *40)
= 44.72
Pn sriram said:
1 decade ago
Awesome answr by chandrasekar.
Kannan said:
1 decade ago
V=(Vr^2+Vc^2)^1/2
V=(400+1600)^1/2
V=45
V=(400+1600)^1/2
V=45
Izhar Rao said:
1 decade ago
Vt=(Vr^2+Vc^2)^1/2
Vt=(20^2+40^2)^1/2
Vt=10(2^2+4^2)^1/2
Vt=10(4+16)^1/2
Vt=10(20)^1/2
Vt=10x4.472
Vt=44.72
Vt=45 volt.
Vt=(20^2+40^2)^1/2
Vt=10(2^2+4^2)^1/2
Vt=10(4+16)^1/2
Vt=10(20)^1/2
Vt=10x4.472
Vt=44.72
Vt=45 volt.
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