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.
Ramesh.H.Adavalli said:
1 decade ago
V/R=R/V
45/3300=R/1200
=16.36V
45/3300=R/1200
=16.36V
Jaime romero said:
1 decade ago
Total resistance= 3300 ohms
Total voltage=45V
R3= 1200 ohms
V3= total voltage*R3/total resistance
V3=45*1200ohms/3300ohms
V3=16.36V
A is the correct answer
Total voltage=45V
R3= 1200 ohms
V3= total voltage*R3/total resistance
V3=45*1200ohms/3300ohms
V3=16.36V
A is the correct answer
Suresh said:
1 decade ago
I = V/R.
= 45/3300 = 9/660,
GIVEN R3 = 1200 Then V3 = ?,
In series combination current is same,
V = 1200*(9/660) = 16.36v.
= 45/3300 = 9/660,
GIVEN R3 = 1200 Then V3 = ?,
In series combination current is same,
V = 1200*(9/660) = 16.36v.
Sai said:
1 decade ago
We know that.
Voltage drop across resistor is (V*R3) /total R.
Such that (45*1200) /3300.
Answer = 16.3636.
Voltage drop across resistor is (V*R3) /total R.
Such that (45*1200) /3300.
Answer = 16.3636.
Shiva said:
10 years ago
Voltage drop mean?
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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