Electronics - Ohm's Law - Discussion
Discussion Forum : Ohm's Law - General Questions (Q.No. 5)
5.
What is the power in the given circuit?
Discussion:
22 comments Page 1 of 3.
Rithin said:
1 decade ago
P=VI , P=I2R, P=V2/R
Nirala anamika kumari jaiprakash said:
1 decade ago
p=(i*i)r
kvl for ckt
7=5r;
therefore r=7/5;
p=(5*5)*7/5=35;
kvl for ckt
7=5r;
therefore r=7/5;
p=(5*5)*7/5=35;
Sundar said:
1 decade ago
i=v/r
p=i*i*r
r=v/r
r=7/5
r=1.4
p=1.4*(5 * 5)
p=35
p=i*i*r
r=v/r
r=7/5
r=1.4
p=1.4*(5 * 5)
p=35
Rohit kumar verma said:
1 decade ago
P=v.i=7.5=35watt
Vimala said:
1 decade ago
Thanks nirala.
Rajesh said:
1 decade ago
power=(I*I)R
We know that R=V/I
=7/5=1.4 ohm
p=(5*5)1.4=35w
We know that R=V/I
=7/5=1.4 ohm
p=(5*5)1.4=35w
BISWAJIT DASH said:
1 decade ago
V=IR
=>R=V/I
=>R=7/5
=>R=1.4
P=I*I*R
=>P=5*5*1.4
=>P=35w
=>R=V/I
=>R=7/5
=>R=1.4
P=I*I*R
=>P=5*5*1.4
=>P=35w
VISWAM said:
1 decade ago
P=VI
V=7
I=5
P=7*5 = 35W
V=7
I=5
P=7*5 = 35W
Sona said:
1 decade ago
Power=Current*Voltage
Ram Ratan Kumar said:
1 decade ago
I=V/R
R=V/I
R=7/5=1.4
P=I*I*R
P=5*5*(1.4)
=35W
R=V/I
R=7/5=1.4
P=I*I*R
P=5*5*(1.4)
=35W
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