Answer to Question #118445 in Electric Circuits for Asif Sultan

Question

http://prntscr.com/so83uo

Answer 1

Explanations & Calculations

• Since the given circuit (1st image) is symmetrical over PQ axis (an axis perpendicular to the input-output axis), voltage distribution in the circuit is also symmetrical.
• Now it could be rearranged to a one shown in the second image (separation along the perpendicular axis).

• Now, resistors in CPD arm are in series; their equivalent is (R+R =2R)
• That 2R and the R above are parallel to each other which make an equivalence of$\Big(\small \frac{2R\times R}{2R+R} =\Big)$ $\frac{2R}{3}$.
• And the resistors between A an B (second image) are in series, equivalent is (R +R =) 2R.

• Now the simplified form is the third image.

• Now resistors between ACDB are in series which make an equivalent of (R +2/3R +R =) $\frac{8}{3}$R.

• Now that resistance & the 2R above are in a parallel arrangement and equivalent to a $\frac{\frac{8R}{3}\times2R}{\frac{8R}{3}+2R}$ = $\frac{8R}{7}$
• Therefore, the equivalent of the circuit is 8/7 R.

• Refer to the equations for the equivalence of both parallel & series arrangements and also see the answer for question Q116210