Electrical Engineering - Alternating Current and Voltage - Discussion
Discussion Forum : Alternating Current and Voltage - General Questions (Q.No. 16)
16.
Two series resistors are connected to an ac source. If there are 7.5 V rms across one resistor and 4.2 V rms across the other, the peak source voltage is
Discussion:
11 comments Page 1 of 2.
HAris said:
6 years ago
Drop accross series resistor = 7.5+4.2 = 11.7Vrms.
Vrms = 0.707*Vp.
Vp = Vrms/0.707.
= 11.7/0.707 = 16.54V.
Vrms = 0.707*Vp.
Vp = Vrms/0.707.
= 11.7/0.707 = 16.54V.
Vinny said:
7 years ago
R1+R2=7.5+4.2=11.7.
Vmax=11.7*1.414=16.543v.
Vmax=11.7*1.414=16.543v.
(1)
EBENESAR said:
7 years ago
Rms value is equal to dc voltage so we can add.
Parth said:
8 years ago
Hypothetically they are in phase.
As phase difference is not given.
Thus, their direct addition is possible.
As phase difference is not given.
Thus, their direct addition is possible.
Pranjal said:
8 years ago
Yes, you are correct. @HIT B.
We cannot add them directly as they are in AC circuit.
We cannot add them directly as they are in AC circuit.
Hit B said:
10 years ago
As this is AC Circuit, then how can we add two voltages algebraically. We have to do complex summation.
Uttam Kumar Sarkar said:
1 decade ago
Vrms = Vmax/1.414
Vmax = Vrms*1.414
= (7.5+4.2)*1.414
= 16.54 Volt
Vmax = Vrms*1.414
= (7.5+4.2)*1.414
= 16.54 Volt
Siddu said:
1 decade ago
Here 7.5/.707 = 10.6
4.2/.707 = 5.800000
10.6 + 5.8 = 16.5
4.2/.707 = 5.800000
10.6 + 5.8 = 16.5
BALBIR KAIRAY said:
1 decade ago
If in the statement component connect in series then component value will be add on all time.
Vinay said:
1 decade ago
peak source voltage= (v1rms+v2rms)*1.414
= (7.5+4.2)*1.414 = 16.54
= (7.5+4.2)*1.414 = 16.54
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers