Electrical Engineering - Alternating Current and Voltage - Discussion
Discussion Forum : 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
resistor is 4 mA, the peak voltage drop across the resistor isDiscussion:
9 comments Page 1 of 1.
Zabi Ullah said:
5 years ago
V= √2*Irms.
ARUN KUMAR M said:
1 decade ago
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.
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.
(4)
Samir kumar sahu said:
1 decade ago
We know V = IR.
So = 4*4.7 = 18.8.
So = 4*4.7 = 18.8.
Arvind kumar prasad said:
1 decade ago
First we can calculate:
v(rms) = 4700*0.004.
= 18.8.
After that we can calculate,
v(max) = 18.8*1.414 = 26.6.
v(rms) = 4700*0.004.
= 18.8.
After that we can calculate,
v(max) = 18.8*1.414 = 26.6.
RAMA CHANDRA RATHA said:
1 decade ago
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.
So peak value = 1.414*rms value.
So peak value of voltage drops = 1.414*18.8 = 26.587 = 26.6.
(2)
Rahul Patil said:
1 decade ago
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
here r.m.s value=(v=ir)4700*0.004=18.8
so peak value=1.414*18.8=26.5832
Rahul Gupta said:
1 decade ago
Peak value = 1.414*4.7*1000*0.4*0.1 = 26.6
Jijo said:
1 decade ago
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
I=4mA=0.004 Amp
V=IxR=0.004*4700=18.8
peak value=(square root 2)* v= 1.414*18.8=26.587
Sunaina said:
1 decade ago
peak value = (square root 2)* v* IR
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