Electronics - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 6)
6.
With a total resistance of 3300 ohms and a 45 V source, if R3 is 1200 ohms, what will be its voltage drop?
Discussion:
16 comments Page 2 of 2.
RAJIESWAR said:
1 decade ago
3300 ohm --->45v.
1200 ohm ----> V2?
V2 = 1200*45/3300.
V2 = 16.3V.
1200 ohm ----> V2?
V2 = 1200*45/3300.
V2 = 16.3V.
(3)
Shiva said:
10 years ago
Voltage drop mean?
Kenneth Bartolo said:
9 years ago
Given : Rtotal = 3300 ohms, V = 45V, R3 = 1200 ohms.
Required: Voltage across R3 = ?
Solution:
Itotal = V / Rtotal = 45/3300.
Itotal = 3/220 A.
Then,
V = IR3.
V = (3/220)(1200),
v = 16.36 V.
Required: Voltage across R3 = ?
Solution:
Itotal = V / Rtotal = 45/3300.
Itotal = 3/220 A.
Then,
V = IR3.
V = (3/220)(1200),
v = 16.36 V.
(1)
Samir ali said:
9 years ago
Use the equation of voltage devider.
vout = vin R3/(R1 + R2) + R3.
R3 = 1200 ohm.
R1 + R2 = Rt - R3.
vout = vin R3/(R1 + R2) + R3.
R3 = 1200 ohm.
R1 + R2 = Rt - R3.
(1)
Mahessh said:
8 years ago
Total resistance=3300ohms,And voltage 45 v.
If R3=1200 ohms and V=?
V=IR.
I=V/R => 45/3300.
I=3/220 A.
V=3/220*1200 =16.36.
V=16.36v.
If R3=1200 ohms and V=?
V=IR.
I=V/R => 45/3300.
I=3/220 A.
V=3/220*1200 =16.36.
V=16.36v.
(5)
Rahul H P said:
5 months ago
Since batteries are connected opposing each other, the current in the circuit will range from a higher voltage to a lower voltage.
It is just like 2 water tanks connected with a single pipe.
The total current in the circuit I = Effective voltage/Total Resistance.
Effective Voltage = V2-V1 (since V2>V1&connected opposite to each other),
Total Resistance = R1 + R2 + R3 + R4 (since all are in series).
I = (3/150)x10^-3
I = 0.02x10^-3 A
So, the current flow is from V2 to V1;
VB = V2-(VR3+VR4) (since the current is equal in the series circuit);
VB = 12 - (0.02 x 10^-3 x 20 x 10^3 + 0.02 x 10^-3 x 56 x 10^3),
VB= 12 - 0.02X76,
VB = 12 - 1.52,
VB = 10.48 V.
VB is Measured only with respect to V2 because there is no current from V1 towards VB so VB is only dropped across R3& R4.
It is just like 2 water tanks connected with a single pipe.
The total current in the circuit I = Effective voltage/Total Resistance.
Effective Voltage = V2-V1 (since V2>V1&connected opposite to each other),
Total Resistance = R1 + R2 + R3 + R4 (since all are in series).
I = (3/150)x10^-3
I = 0.02x10^-3 A
So, the current flow is from V2 to V1;
VB = V2-(VR3+VR4) (since the current is equal in the series circuit);
VB = 12 - (0.02 x 10^-3 x 20 x 10^3 + 0.02 x 10^-3 x 56 x 10^3),
VB= 12 - 0.02X76,
VB = 12 - 1.52,
VB = 10.48 V.
VB is Measured only with respect to V2 because there is no current from V1 towards VB so VB is only dropped across R3& R4.
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