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Current Electricity · Example 3.32 (i)

Equivalent Resistance Between P and Q

Find the equivalent resistance between P and Q for the symmetric bridge network shown (note: r and R denote the same value).

P Q T X U 4R 4R 2r 2r 4R 4R R R
Top/bottom arms 4R; middle arms 2r; the two central rungs (T–X and X–U) are R.
Symmetry note: T, X and U all lie on the perpendicular bisector of PQ, so they are at the same potential ⇒ no current through the two R rungs ⇒ delete them.
⬇ drop the zero-current R's
P Q 4R4R 2R2R 4R4R = 8R = 4R = 8R
Three independent parallel branches: 8R, 4R, 8R.
Given
Asked

Equivalent resistance RPQ.

Key Idea + Formula

On the perpendicular bisector of PQ every point is at potential \(\tfrac{V_P+V_Q}{2}\). Equal potential ⇒ zero current ⇒ that resistor is removable. Then:

\[ \frac{1}{R_{PQ}} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \]
Solution
STEP 1 — Kill the central rungs
T, X, U sit on the symmetry axis ⇒ \(V_T = V_X = V_U\) ⇒ no current through either R ⇒ remove both.
STEP 2 — Three clean series branches remain
Top: \(4R+4R = 8R\);   Middle: \(2r+2r = 4R\);   Bottom: \(4R+4R = 8R\).
STEP 3 — Parallel them
\[ \frac{1}{R_{PQ}} = \frac{1}{8R}+\frac{1}{4R}+\frac{1}{8R} = \frac{1+2+1}{8R} = \frac{1}{2R} \] \[ R_{PQ} = 2R \]
Equivalent Resistance RPQ = 2R
Short Cut

Any rung lying on the line of symmetry carries no current. When the network mirrors across the perpendicular bisector of the input terminals, the bridge/cross resistors sitting on that axis are dead — erase them first, and what's left is almost always plain parallel branches.

Finish fast: 8R ∥ 8R = 4R, then 4R ∥ 4R = 2R (equal pair halves). This is the same network the next page redraws as 8R, 4R, 8R in parallel.

Given → Asked → Formula → Solution → Short Cut · verified by nodal analysis (R = 2R, r = R)