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Electrical Engineering - Alternating Current and Voltage - Discussion

Discussion Forum : Alternating Current and Voltage - General Questions (Q.No. 1)
Alternating Current and Voltage
1.
If the rms voltage drop across a 15 k resistor is 16 V, the peak current through the resistor is
15 mA
1.5 mA
10 mA
1 mA
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
27 comments Page 1 of 3.
Naveen said:   1 decade ago
V= 16
R= 15oo0
Irms= 16/15000
Im= Irms/0.707
= 16/(15000*0.707)
= 1.5 mA
ROKON said:   1 decade ago
(RMS x 1.414) x 2=P-P
Nafish said:   1 decade ago
It as simple as.....
V=16
R=15k = 15000
Irms= V/R = 16/15000 = 0.001067A = 1.067mA
Imax=1.414*1.067mA = 1.51mA
Keerthi said:   1 decade ago
vrms=vm/1.414
so given vrms=16
hence vm=16*1.414
vm=22.6 v
im=vm/r
im=22.6/15k
im=1.5mA
Rakesh kumar said:   1 decade ago
im=2^1/2*(irms)
irms=v/r
=16/15k
im=2^1/2*(16/15k)
=1.51ma
Yuvraj said:   1 decade ago
Irms=Vrms/15k=1.066 ma
now Ipeak=Irms*sqrt(2)

therefore, Ipeak= 1.066*1.414 =1.5 ma
Ayush said:   1 decade ago
V=16
R=15k = 15000
Irms= V/R = 16/15000 = 0.001067A = 1.067mA
Imax=1.414*1.067mA = 1.51mA
Rajesh said:   1 decade ago
Here giving,

V = 15 v.
And R = 15000.

Then we know that Im = 2^1/2 Irms.

And Irms = V/R and solving it find Im = 1.5 mA.
DHARMA said:   1 decade ago
Mr @Rajesh said that v=15v. Then how he calculated and answer also 1. 5mA. HOW?
Lakmal said:   1 decade ago
Please Anyone answer me. In which level This questions are include? In Diploma level?
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