Electrical Engineering - Alternating Current and Voltage - Discussion

Discussion Forum : Alternating Current and Voltage - General Questions (Q.No. 1)
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 2 of 3.

Aamir sohail said:   1 decade ago
Sample,

Imax = Irms*1.414
Irms = V/R
V=16, R=15 k=15000.

Irms = 16/15000 = 0.001067A = 1.067mA.
Imax = 1.067*1.414 = 1.5mA.

GANESH said:   1 decade ago
Irms = Io/1.414.
Io = Eo/R=16/15mA.
Irms = 16*0.707/15mA = 1.5mA.

Gopi said:   1 decade ago
Erms = 16v.
R = 15Kohms = 15*10^3 = 15000.
We have Erms = Irms*R.
Irms = Erms/R.
= 16/15000.
= 0.001066A.

We have Irms = Imax/1.414.

Imax = Irms*1.414.
= 0.001066*1.414.
= 0.0015A = 1.5mA.

Dino said:   1 decade ago
Emax = 16 V/0.707.

Emax = 22.63 V or 22V.

I = V/R.

I = 22 V/ 15 kΩ.

to convert 15 kΩ to ohms.

15 kΩ * 1000 Ω/1 kΩ.

I= 22 V/ 15000 Ω.

I= 1.46 mA or 1.5 mA.

Nikita said:   1 decade ago
Vpeak = 1.414 * Vrms = 1.414*16 = 22.624.

Ipeak = Vpeak/R = 22.624/15000 = 1.5mA.

Sunil kumar said:   1 decade ago
rms voltage = 16 volts.

max voltage = 16*1.414= 22.62 volts.

I=V/R =22.62/15000 = 1.5 mA.

GANGA RAM GAUTAM said:   1 decade ago
V = 15V.
R = 15000.

Irms = V/R = 16/15000.
= 0.001067A.

Imax = 1.414x1.067mA.
= 1.51mA.

Lakmal said:   1 decade ago
Please Anyone answer me. In which level This questions are include? In Diploma level?

DHARMA said:   1 decade ago
Mr @Rajesh said that v=15v. Then how he calculated and answer also 1. 5mA. HOW?

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.


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