Hi, everybody!
The question here is how to find the Thevenin equivalent circuit left of the load \( R_l \).
\(
k =10
a) R_2 = \infty
b) R_2 = 50 k \Omega
\)
Applying KCL to nodes C and A of the circuit which is open at A-B:
\(
\frac{10 - V_d}{10000} + \frac{v_A - v _d}{R_2}
\frac{v_d - v_A}{R_2} + \frac{-k v_d - v_A}{100}
\)
Solving these equations we get:
a) \( V_{Th} = V_a = - 100 V\)
b) \( V_{Th} = V_a = - 31.224 V\)
To find the Thevenin resistance, the circuit is shorted at A-B and all the voltage sources:
\( R_{Th} = (10k \Omega + R_2) \parallel (100 \Omega) \)
a) \( R_{Th} = 100 \Omega \)
b) \( R_{Th} = 99.834 \Omega \)
According to the textbook, these results are correct except \( R_{Th} \) in (b), which is given there as \( 37.48 \Omega \). So in (a), the circuit provides \( - 1A \) current to the load resistor. Using the result from (b) here, the current should be \( -0.313 A \), while according to the textbook the current should be \( -0.833 A \). So according to the textbook, changing \( R_2 \) from \( \infty \) to \( 50 k \Omega \) reduces the output current by only \( \approx 17% \). Which solution to \( R_{Th} \) in (b) is then correct? Should not using the same formula for both Thevenin resistances in (a) and (b) guarantee that if one is correct that the other is too?
The question here is how to find the Thevenin equivalent circuit left of the load \( R_l \).
\(
k =10
a) R_2 = \infty
b) R_2 = 50 k \Omega
\)
Applying KCL to nodes C and A of the circuit which is open at A-B:
\(
\frac{10 - V_d}{10000} + \frac{v_A - v _d}{R_2}
\frac{v_d - v_A}{R_2} + \frac{-k v_d - v_A}{100}
\)
Solving these equations we get:
a) \( V_{Th} = V_a = - 100 V\)
b) \( V_{Th} = V_a = - 31.224 V\)
To find the Thevenin resistance, the circuit is shorted at A-B and all the voltage sources:
\( R_{Th} = (10k \Omega + R_2) \parallel (100 \Omega) \)
a) \( R_{Th} = 100 \Omega \)
b) \( R_{Th} = 99.834 \Omega \)
According to the textbook, these results are correct except \( R_{Th} \) in (b), which is given there as \( 37.48 \Omega \). So in (a), the circuit provides \( - 1A \) current to the load resistor. Using the result from (b) here, the current should be \( -0.313 A \), while according to the textbook the current should be \( -0.833 A \). So according to the textbook, changing \( R_2 \) from \( \infty \) to \( 50 k \Omega \) reduces the output current by only \( \approx 17% \). Which solution to \( R_{Th} \) in (b) is then correct? Should not using the same formula for both Thevenin resistances in (a) and (b) guarantee that if one is correct that the other is too?