# Compute voltages and equivalent resistance in the presence of a bridge

Problem. Given the circuit in the figure, I wish to compute the equivalent resistance between node A and D, voltage $$\V_{\mathrm{BC}}\$$, both under closed and open bridge K. $$\R_1=50\,\Omega\$$, $$\R_2=40\,\Omega\$$, $$\R_3=60\,\Omega\$$,$$\R_4=80\,\Omega\$$, $$\R_5=30\,\Omega\$$, $$\V_{\mathrm{AD}}=50\$$ V.

Solution. With K open, all is fine, I get $$\R_{eq} = 45.5\$$ and $$\V_{BC} = 9.34\$$ V.

K closed. Considering the star-type connection $$\R_2,R_3,R_5\$$. I transform it to an equivalent triangle and then get the equivalent resistance as follows (picture below):

\begin{align*} R_{\mathrm {EB}} &= \frac{R_2R_3+R_2R_5+R_3R_5}{R_5}=\frac{40\cdot60+40\cdot30+60\cdot30}{30}= 180\,\Omega\\ R_{\mathrm {BF}} &= \frac{R_2R_3+R_2R_5+R_3R_5}{R_2}= 135\,\Omega,\quad R_{\mathrm {EF}} = \frac{R_2R_3+R_2R_5+R_3R_5}{R_3}= 90\,\Omega. \end{align*}

Furthermore, \begin{align*} R_{\mathrm{1EB}} &= \frac{1}{R_1^{-1}+R_{\mathrm {EB}}^{-1}} = 39.13\,\Omega,\quad R_{\mathrm{4BF}} = \frac{1}{R_4^{-1}+R_{\mathrm {BF}}^{-1}} = 50.23\,\Omega\\ R_{\mathrm{E14F}} &= R_{\mathrm{1EB}}+R_{\mathrm{4BF}} = 89.36\,\Omega,\quad R_{\mathrm{eq}} = \frac{1}{R_{\mathrm{E14F}}^{-1}+R_{\mathrm {EF}}^{-1}} = 44.8\,\Omega. \end{align*} From the figure below, the total current is: $$\I = V_{\mathrm{AD}}/R_{\mathrm{eq}} = 1.116\$$ A. After node E, this current divides into two \begin{align*} I & =I_{14} + I_{\mathrm{EF}},\quad I_{\mathrm{EF}} = V_{\mathrm{AD}}/R_{\mathrm{EF}} = 0.556\,A,\quad I_{14} = V_{\mathrm{AD}}/(R_{\mathrm{1EB}} +R_{\mathrm{4BF}}) = 0.56\,A. \end{align*}

In node B, there is a drop of voltage due to resistor $$\R_{\mathrm{1EB}}\$$, thus $$\V_{\mathrm{B}} = I_{14}R_{\mathrm{1EB}} = 21.9\$$ hence the potential $$\V_{\mathrm{BA}}=50-21.9=28.1\$$ V.

Now consider the star $$\R_1,R_3,R_4\$$ (picture below). We have

\begin{align*} R_{\mathrm {GC}} &= \frac{R_1R_3+R_1R_4+R_3R_4}{R_4}=\frac{50\cdot60+50\cdot80+60\cdot80}{80}= 147.5\,\Omega\\ R_{\mathrm {GH}} &= \frac{R_1R_3+R_1R_4+R_3R_4}{R_3}= 196.7\,\Omega,\quad R_{\mathrm {CH}} = \frac{R_1R_3+R_1R_4+R_3R_4}{R_1}= 236\,\Omega. \end{align*}

Hence \begin{align*} R_{\mathrm{2GC}} &= \frac{1}{R_2^{-1}+R_{\mathrm {GC}}^{-1}} = 33.24\,\Omega,\quad R_{\mathrm{5CH}} = \frac{1}{R_5^{-1}+R_{\mathrm {CH}}^{-1}} = 26.62\,\Omega\\ R_{\mathrm{E25F}} &= R_{\mathrm{2GC}}+R_{\mathrm{5CH}} = 59.86\,\Omega,\quad R_{\mathrm{eq}}^{'} = \frac{1}{R_{\mathrm{E25F}}^{-1}+R_{\mathrm {GH}}^{-1}} = 45.89\,\Omega. \end{align*}

To get the drop of voltage at node C note that the current is `e $$\I = V_{\mathrm{AD}}/R_{\mathrm{eq}}^{'} = 1.09\$$ A, which, afer node E divides as

\begin{align*} I & =I_{24} + I_{\mathrm{GH}},\quad I_{\mathrm{GH}} = V_{\mathrm{AD}}/R_{\mathrm{GH}} = 0.254\,A,\quad I_{14} = V_{\mathrm{AD}}/(R_{\mathrm{2GC}} +R_{\mathrm{5CH}}) = 0.836\,A. \end{align*}

The drop of voltage in C, due to resistor $$\R_{\mathrm{2GC}}\$$ is $$\V_{\mathrm{C}} = I_{24}R_{\mathrm{2GC}} = 8.4\$$ so the potential is $$\V_{\mathrm{CA}}=50-8.4=41.6\$$ V.

The difference in potential between B and C is $$\V_C -V_B= 13.5\$$ V.

Comments As you can see I get different equivalent resistances, i.e. $$\R_{eq}\neq R_{eq}^{'}\$$, which is impossible. Furthermore my potential $$\V_{BC}\$$ is too high (ans. 5.19 V).

Question What am I missing?

• IMHO you've made the calculations more complicated to follow by introducing unnecessary extra node letters since A,E&G are the same node as are D,F&H. So, for example, Rgc could be Rac which is in parallel with R2 Commented Sep 10, 2022 at 8:39
• Yes, there is some redundancy in the figure, but for the sole purpose to fix my ideas (I'm a newbie!). Commented Sep 10, 2022 at 9:19

I've just checked the first half of the problem by transforming the delta R3, R4, R5 to a star and get the equivalent resistance to be 44.8 ohms. Which agrees with your first set of calculations. Therefore I suggest that there is some error in your calculations for the second version.

In your second set of calculations your star-delta transform of R1, R3, R4 is correct, but when you calculate R2GC , 147.5 in parallel with 40 you should get 31.47 not 33.24. If you use this value in the rest of the calculation you get the same equivalent resistance, 44.8

• thank you! What about the star R1, R3, R4? What do you obtain? Commented Sep 9, 2022 at 22:47
• I've found your mistake and will update my answer. Commented Sep 9, 2022 at 23:03

Nodal analysis on the original circuit gives,

$$\\large \frac{V_B-50}{R_1}+\frac{V_B-V_C}{R_3}+\frac{V_B}{R_4}\small=0 \$$

$$\\large \frac{V_C-50}{R_2}+\frac{V_C-V_B}{R_3}+\frac{V_C}{R_5}\small= 0 \$$

Solve by simultaneous equations,

$$\\small V_B =28.106\:V\$$

$$\\small V_C =22.912\:V\$$

$$\\small V_B-V_C =5.193\:V\$$

• @utobi When you write a comment, the input box literally displays this warning: Avoid comments like "+1" or "thanks".
– pipe
Commented Sep 10, 2022 at 9:37