INTRODUCTION TO ELECTRONICS: The Superposition Theorem

Q: How much current flows across R₂? What is the voltage drop across R₂?

A: We begin by evaluating the circuit from two points of view: V_s1 and V_s2. In order to begin our journey, the V_s2 voltage source is shorted:

From the vantage point of V_s1, resistor R₁ is in series with resistors R₂ and R₃, which are parallel to one another:

R_T(s1) = R₁ + R₂∥R₃

R_T(s1) = 100 Ω + [ ( 220 Ω )( 330 Ω ) / ( 220 Ω + 330 Ω ) ]

R_T(s1) = 100 Ω + [ ( 220 Ω )( 330 Ω ) / ( 550 Ω ) ]

R_T(s1) = 100 Ω + 132 Ω

R_T(s1) = 232 Ω

I_T(s1) = ( 10 V / 232 Ω )

I_T(s1) = 43.1 mA

I_2(s1) = [ R₃ / ( R₂ + R₃ ) ]( I_T(s1) )

I_2(s1) = [ 330 Ω / ( 550 Ω ) ]( 43.1 mA )

I_2(s1) = ( 0.6 )( 43.1 mA )

I_2(s1) = 25.9 mA

The same procedure is now used to evaluate the circuit from the V_s2 vantage point:

R_T(s2) = R₃ + R₁∥R₂

R_T(s2) = 330 Ω + [ ( 100 Ω )( 220 Ω ) / ( 320 Ω ) ]

R_T(s2) = 330 Ω + 68.8 Ω

R_T(s2) = 399 Ω

I_T(s2)= ( V_(s2)/ R_T(s2))

I_T(s2)= ( 5 V / 399 Ω )

I_T(s2)= 12.5 mA

I_2(s2)= [ R₁ / ( R₁ + R₂ ) ]( I_T(s2) )

I_2(s2) = [ 100 Ω / ( 320 Ω ) ]( 12.5 mA )

I_2(s2) = ( 0.31 )( 12.5 mA )

I_2(s2) = 3.88 mA

I_2T = I_2(s1) + I_2(s2)

I_2T = 25.9 mA + 3.88 mA

I_2T = 29.8 mA

V_R2 = ( I_2T )( R₂ )

V_R2 = ( 29.8 mA )( 220 Ω )

V_R2 = 6.56 V

Published by George Tafari