1. Three resistors in series have an Rt of 7kΩ. If R3 is 2.2 times larger than R1 and 1.5 times larger than R2, what are the values of R1, R2, and R3?
Answers
Answer:
R₁ = 1.5kΩ
R₂ = 2.2kΩ
R₃=3.3kΩ
Step-by-step explanation:
Resistance of combination of resistors in series Rs = R₁+R₂+R₃. Let this be equation 1.
R₃=2.2 × R₁. Let this be equation 2.
R₃=1.5 × R₂. Let this be equation 3.
Now, equation 2 = equation 3.
∴2.2 × R₁ = 1.5 × R₂
R₂ = (2.2 × R₁) / 1.5. Let this be equation 4.
Now, let us substitute eqn. 4 and eqn. 2 in eqn. 1
Rs = 7kΩ = 7000Ω
∴7000 = R₁+(2.2 × R₁) / 1.5 + 2.2 × R₁
7000 = ( [1.5 × R₁] + [2.2 × R₁] + [2.2 × 1.5 × R₁] ) / 1.5
10500 = ( [1.5 × R₁] + [2.2 × R₁] + [3.3 × R₁] )
10500 = 7 × R₁
∴ R₁ = 10500 / 7
R₁ = 1500Ω or R₁ = 1.5kΩ
Now, Substituting R₁ in eqn. 4, we can obtain R₂
R₂ = (2.2 × R₁) / 1.5
R₂ = (2.2 × 1500) / 1.5
R₂ = 2200Ω or R₂ = 2.2kΩ
Next, let us substitute R₂ in eqn. 3 to obtain R₃.
R₃=1.5 × R₂
R₃=1.5 × 2200
R₃=3300Ω or R₃=3.3kΩ
VERIFICATION:
Rs = R₁+R₂+R₃
Rs = (1.5 + 2.2 + 3.3) kΩ
Rs = 7kΩ
Hence, verified.