Multiloop Circuits — Question 5

Question 5

A two-source network can be replaced by its open-circuit voltage and resistance

Two branches connect ground to output V: an upward 12.0 V ideal source in series with 3.00 ohm, and an upward 6.00 V source in series with 6.00 ohm. Define a Thevenin equivalent as an ideal voltage Vth in series with Rth.

1. Find Vth with the output open and Rth with independent ideal voltage sources replaced by shorts.

2. Attach a 3.00 ohm load and find its current and voltage. Verify by the original node equation.

Question 5 - Solution

A two-source network can be replaced by its open-circuit voltage and resistance

(a) Method and calculation

$$(12-V_{\mathrm{th}})/3+(6-V_{\mathrm{th}})/6=0\Rightarrow V_{\mathrm{th}}=10.0\,\mathrm V.$$$$R_{\mathrm{th}}=3\parallel6=2.00\,\Omega.$$

(b) Interpretation and further calculation

$$I_L=\frac{V_{\mathrm{th}}}{R_{\mathrm{th}}+R_L}=10/5=2.00\,\mathrm A,\quad V=I_LR_L=6.00\,\mathrm V.$$$$(12-6)/3+(6-6)/6=2.00\,\mathrm A=V/R_L.$$

Physical check

Shorting an ideal voltage source means setting its emf to zero while retaining all branch resistors.