Electrical Engineering - Alternating Current and Voltage - Discussion

Discussion :: Alternating Current and Voltage - General Questions (Q.No.6)

6. 

If the rms current through a 4.7 k resistor is 4 mA, the peak voltage drop across the resistor is

[A]. 4 V
[B]. 18.8 V
[C]. 26.6 V
[D]. 2.66 V

Answer: Option C

Explanation:

No answer description available for this question.

Sunaina said: (Jan 12, 2011)  
peak value = (square root 2)* v* IR

Jijo said: (Jul 20, 2011)  
R =4.7kilo ohms=4700 ohms
I=4mA=0.004 Amp
V=IxR=0.004*4700=18.8

peak value=(square root 2)* v= 1.414*18.8=26.587

Rahul Gupta said: (Oct 30, 2011)  
Peak value = 1.414*4.7*1000*0.4*0.1 = 26.6

Rahul Patil said: (Dec 25, 2011)  
peak value=(square root 2 )*r.m.s value
here r.m.s value=(v=ir)4700*0.004=18.8
so peak value=1.414*18.8=26.5832

Rama Chandra Ratha said: (Sep 30, 2012)  
Here rms value of voltage drop = i*r = 0.004*4700 = 18.8.

So peak value = 1.414*rms value.

So peak value of voltage drops = 1.414*18.8 = 26.587 = 26.6.

Arvind Kumar Prasad said: (Mar 26, 2014)  
First we can calculate:
v(rms) = 4700*0.004.
= 18.8.

After that we can calculate,
v(max) = 18.8*1.414 = 26.6.

Samir Kumar Sahu said: (Nov 15, 2014)  
We know V = IR.

So = 4*4.7 = 18.8.

Arun Kumar M said: (Aug 22, 2015)  
Im = Irms/0.707.
Vm = Vrms 1.414.

V = IR Vrms = 4*4.7 = 18.8.
Vm = Vrms*1.414.
Vm = 18.8*1.414.

Peak voltage Vm = 26.58.

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