Electronics - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 11)
11.
An 8-ohm resistor is in series with a lamp. The circuit current is 1 A. With 20 V applied, what voltage is being allowed for the lamp?
Discussion:
5 comments Page 1 of 1.
Sreeyush Sudhakaran said:
1 decade ago
V(Lamp)+V(Resistor)=V(Total) (Total Potential=Sum of all potential drops)
V(Resistor)=IxR
I=1A R=8ohms V(Total)=20V
V(Resistor)=1x8=8V
V(Lamp)=V(Total)-V(Lamp)
V(Lamp) = 20V-8V = 12V
V(Resistor)=IxR
I=1A R=8ohms V(Total)=20V
V(Resistor)=1x8=8V
V(Lamp)=V(Total)-V(Lamp)
V(Lamp) = 20V-8V = 12V
(1)
Kiran Hatti said:
1 decade ago
The voltage drop across resistor 8ohm id V=IxR=8V. So the voltage across lamp is total voltage minus voltage across resistor ie, VL=VT-VR=12V
Manoj khandare said:
10 years ago
First find voltage v = i*r = 8*1.
Now applied voltage is 20 v.
So lamp voltage = 20-8 = 12 v.
Now applied voltage is 20 v.
So lamp voltage = 20-8 = 12 v.
(2)
M.PANDIYARAJAN said:
1 decade ago
V=I*R
V=1*8
V=8
Applied voltage=20
Being voltage=20-8
=12v
V=1*8
V=8
Applied voltage=20
Being voltage=20-8
=12v
Avinash said:
1 decade ago
20 = 8 + lamp voltage.
Lamp voltage = 20-8.
Lamp voltage = 12.
Lamp voltage = 20-8.
Lamp voltage = 12.
(1)
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