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

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.