Electronics - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 18)
18.
What is the total power in the circuit?
Discussion:
4 comments Page 1 of 1.
Sahil Chaudhari said:
5 years ago
P = V^2/R
R = R1 + R2 + R3 + R4 = 13.2kΩ.
P = (48)^2/13.2k = 174.54mΩ.
=> 170mΩ.
R = R1 + R2 + R3 + R4 = 13.2kΩ.
P = (48)^2/13.2k = 174.54mΩ.
=> 170mΩ.
Sekhar said:
1 decade ago
Total resistance(R) = 13.2K.
current I = V/R = 48/13.2K = 3.636mA.
Power = I^2xR.
P = (1.322x10^-5).(13.2K).
= 174mW = 170mW.
current I = V/R = 48/13.2K = 3.636mA.
Power = I^2xR.
P = (1.322x10^-5).(13.2K).
= 174mW = 170mW.
Venki said:
1 decade ago
P=VI
I=V/R
I=48/(1+1.5+3.3+7.4)
=48/13.2k ohms
=48/13200
=0.00363
Then p=48*0.00363
=0.174
=170mw
I=V/R
I=48/(1+1.5+3.3+7.4)
=48/13.2k ohms
=48/13200
=0.00363
Then p=48*0.00363
=0.174
=170mw
(1)
Naseem ul aziz said:
1 decade ago
p=vi
i=v/r
p=v^2/r
v=48 volt
r=13200 Ohm
p=2304/13200=0.170=170 mw
i=v/r
p=v^2/r
v=48 volt
r=13200 Ohm
p=2304/13200=0.170=170 mw
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