Please help me solve this problem
6) The following circuits are both voltage dividers. a) Solve for the value of R1 in the following circuit: b) Solve for Va and Vb in the following circuit (referenced to ground as
Solution
a) Given a voltage source of 10Volts.
We can clearly see the voltage is being DIVIDED between R1 and R2.
So apply voltage division rule here:
And the voltage at R2 is indicated as 2.5Volts, Now cleaverly use this in the formula
2.5Volts = (10Volts X 2K)/(2K + R1)
2K + R1 = 8K
R1 = 8k - 2k
R1 = 6K
b) Now apply KVL technique for this circuit
Single mesh we have, so we get a single equation
6 = i(1K) + i(2k)
6 = i(3k)
i = 6 / 3k
i = 2mA
So the voltage Va is given by i X R1
Va = 2mA X 1K
Va = 2Volts
And Vb = - i X 2K
Vb = - 2mA X 2K
Vb = - 4 Volts
.
Please help me solve this problem 6) The following circuits are both v.docx
1. Please help me solve this problem
6) The following circuits are both voltage dividers. a) Solve for the value of R1 in the following
circuit: b) Solve for Va and Vb in the following circuit (referenced to ground as
Solution
a) Given a voltage source of 10Volts.
We can clearly see the voltage is being DIVIDED between R1 and R2.
So apply voltage division rule here:
And the voltage at R2 is indicated as 2.5Volts, Now cleaverly use this in the formula
2.5Volts = (10Volts X 2K)/(2K + R1)
2K + R1 = 8K
R1 = 8k - 2k
R1 = 6K
b) Now apply KVL technique for this circuit
Single mesh we have, so we get a single equation
6 = i(1K) + i(2k)
6 = i(3k)
i = 6 / 3k
i = 2mA
So the voltage Va is given by i X R1
2. Va = 2mA X 1K
Va = 2Volts
And Vb = - i X 2K
Vb = - 2mA X 2K
Vb = - 4 Volts
