Q.A battery of $10\ \text{V}$ and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance $1\ \Omega$ (Fig. 3.16). Determine the equivalent resistance of the network and the current along each edge of the cube.

The key insight is that the three resistors are identical and form a closed loop (a delta network). The 3 A current enters at vertex P and leaves at vertex R. Because the resistors are all $1\,\Omega$, the network is symmetric with respect to the line through P and the midpoint of QR — but here the exit is at R, not at the midpoint, so we cannot simply split the current equally. Instead, we must apply Kirchhoff’s current law (KCL) at the junctions and Kirchhoff’s voltage law (KVL) around the loop.

Let’s label the currents clearly:

• $i_1$ flows from P to R (right side).
• $i_2$ flows from P to Q (left side).
• Let $i_3$ be the current from Q to R along the bottom side.

At vertex P, the incoming 3 A splits into $i_1$ and $i_2$:

At vertex Q, current $i_2$ arrives from P, and then leaves via the bottom resistor toward R. So at Q:

At vertex R, current $i_1$ arrives from P, and current $i_3$ arrives from Q; both leave the network as the outgoing current (which must be 3 A, since current is conserved). So:

Now apply Kirchhoff’s voltage law around the triangle P→Q→R→P. Going clockwise:

• From P to Q: voltage drop = $i_2 \times 1\,\Omega = i_2$ (positive if we go with the current direction).
• From Q to R: voltage drop = $i_3 \times 1\,\Omega = i_3$ (direction from Q to R matches our assumed current).
• From R back to P: voltage drop = $-i_1 \times 1\,\Omega = -i_1$ (because we are going opposite to the direction of $i_1$).

Sum of voltage drops around the closed loop must be zero:

Now we have four equations, but only three unknowns. Solve them:

From (2): $i_3 = i_2$.

Substitute into (1) and (3):

• (1): $i_1 + i_2 = 3$
• (3): $i_1 + i_2 = 3$ (same equation, so it’s consistent). Now (4): $i_2 + i_2 - i_1 = 0 \Rightarrow 2i_2 - i_1 = 0 \Rightarrow i_1 = 2i_2$.

Plug into (1): $2i_2 + i_2 = 3 \Rightarrow 3i_2 = 3 \Rightarrow i_2 = 1$ A.

Thus the current through the left side PQ is 1 ampere.

Step 1 — First junction

The figure shows $5\,\text{A}$ entering the first junction, with $1\,\text{A}$ branching off downward. By KCL, the current continuing along the main line is

Step 2 — Second junction

At the second junction, the figure shows $1.5\,\text{A}$ branching off downward from the $I_1 = 4\,\text{A}$ arriving there. By KCL, the current continuing along the main line is

Step 3 — Third junction

At the final junction, the $I_2 = 2.5\,\text{A}$ arriving splits into two branches, one of which the figure marks as $1.5\,\text{A}$. By KCL, the other branch is

At every junction, the current entering equals the sum of the currents leaving, so the full $5\,\text{A}$ supplied at the input is accounted for across the whole network.

Step 1 — Derive the equivalent emf.

For two cells in parallel, the total current is the sum of branch currents. With terminal voltage $V$:

Rearranging into the form $V = E_{eq} - I\,r_{eq}$ gives the standard results

Step 2 — Note that it is a weighted average.

Write it as

A convex combination of $E_1$ and $E_2$ must lie strictly between them, so with $E_2 > E_1$:

This already eliminates (B) and (C).

Step 3 — Decide which emf it is nearer.

Crucially, $E_1$ carries the weight $w_1 = r_2/(r_1+r_2)$ — the other cell's resistance. Given $r_2 > r_1$:

The dominant weight sits on $E_1$, so $E_{eq}$ is nearer to $E_1$.

Sanity check with numbers: $E_1 = 2\,\mathrm{V},\ E_2 = 4\,\mathrm{V},\ r_1 = 1\,\Omega,\ r_2 = 3\,\Omega$:

which indeed lies between 2 V and 4 V, and is much closer to $E_1 = 2$ V. ✓

ANSWER: D

The square ABCD (total wire $4\,\Omega$) has each side $1\,\Omega$: $AB=BC=CD=DA=1\,\Omega$. Wire $AP=1\,\Omega$ joins A to P on DC. Let $DP=x$, $PC=1-x$. Apply p.d. between A and C; take $V_A=V$, $V_C=0$.

Potential of B (branch A–B–C, series $2\,\Omega$): $V_B=\dfrac{V}{2}$.

Impose $V_P=\dfrac{V}{2}$ and use node equations at D and P.

At D: $\dfrac{V_D-V}{1}+\dfrac{V_D-V_P}{x}=0\Rightarrow V_D=\dfrac{V(x+\tfrac12)}{x+1}$.

At P: $\dfrac{V_P-V_D}{x}+\dfrac{V_P}{1-x}+\dfrac{V_P-V}{1}=0$. Substituting (with $V/2-V_D=-\dfrac{Vx}{2(x+1)}$):

Hence $\dfrac{2x}{1-x^{2}}=1\Rightarrow x^{2}+2x-1=0\Rightarrow x=\sqrt{2}-1\ \Omega$.

1. The self-similarity trick

Let $R$ be the resistance looking into the ladder at $A$–$B$. Strip off the first section: a $2\,\Omega$ resistor in the top rail, a $2\,\Omega$ resistor in the bottom rail, and the first $2\,\Omega$ shunt rung. What lies beyond that rung is again an infinite ladder identical to the original, so its resistance is also $R$.

Hence the first shunt ($2\,\Omega$) is in parallel with $R$, and that combination is in series with the two $2\,\Omega$ rail resistors:

2. Solve the equation

Multiply through by $(2+R)$:

3. Choose the physical root

Resistance must be positive, and $2 - 2\sqrt3 \approx -1.46\,\Omega$ is negative, so we discard it:

Sanity check: $R$ must exceed the $4\,\Omega$ of the two series rail resistors alone (the rest of the ladder can only add), and it must be less than $4 + 2 = 6\,\Omega$ (the case where the ladder beyond the first rung were an open circuit... more precisely the shunt alone caps the extra at $2\,\Omega$). Indeed $4 < 5.46 < 6$. ✓

ANSWER: A

The key to solving a cubical resistor network is symmetry. When current enters at A and leaves at B, the geometry of the cube creates points that are at the same potential. You can then “short” those points (connect them without changing the circuit) and redraw the network into a much simpler one.

Let’s work through it step by step.

1. Identify the symmetry.

   The cube has 12 edges, each a $1 \, \Omega$ resistor. Current enters at vertex A and exits at vertex B. Because the cube is symmetric about the line AB, the three vertices that are one edge away from A (call them C, D, E) are all at the same potential. Similarly, the three vertices that are one edge away from B (call them F, G, H) are also at the same potential. The remaining two vertices (the ones diagonally opposite on the faces not containing A or B) are also symmetric.

   Tip

   In a cube, the vertices that are equidistant from the entry and exit points are equipotential. Here, the three neighbours of A are all at one potential, and the three neighbours of B are at another.

2. Redraw the network using equipotential points.

   Since the three neighbours of A are at the same potential, you can join them into a single node. Same for the three neighbours of B. This collapses the cube into a simpler circuit:

   • From A, three $1 \, \Omega$ resistors go to the “middle” node (the three neighbours of A).
   • From that middle node, six resistors go to the other middle node (the three neighbours of B) — but careful: each of the three neighbours of A is connected to two of the neighbours of B via the cube’s edges. That gives $3 \times 2 = 6$ resistors, each $1 \, \Omega$, all in parallel between the two middle nodes.
   • Finally, from the second middle node, three $1 \, \Omega$ resistors go to B.

3. Calculate the equivalent resistance step by step.

   • From A to the first middle node: three $1 \, \Omega$ resistors in parallel.

• Between the two middle nodes: six $1 \, \Omega$ resistors in parallel.

• From the second middle node to B: again three $1 \, \Omega$ resistors in parallel.

These three resistances are in series (current must go from A to the first middle node, then across to the second, then to B). So:

4. Simplify the sum.

In an AC circuit the ammeter and voltmeter measure RMS values. The resistor and inductor are in series across the $220\ \text{V}$, $50\ \text{Hz}$ supply. The ammeter gives the total RMS current, and the voltmeter reads $220\ \text{V}$.

1. Inductive reactance.

With $L = \dfrac{0.7}{\pi}\ \text{mH}$ and $f = 50\ \text{Hz}$,

This is negligible compared with $R = 100\ \Omega$, so the impedance is essentially resistive: $Z \approx R = 100\ \Omega$.

2. Ammeter reading (RMS current).

3. Voltmeter reading.

The voltmeter reads the supply voltage, $220\ \text{V}$.

Thus the readings are $2.2\ \text{A}$ and $220\ \text{V}$.

Step 1 — Map the network onto its nodes

From the figure the connections are:

• $P \to M$ : $4\,\Omega$
• $M \to R$ (apex) : $3\,\Omega$
• $R \to N$ : $3\,\Omega$
• $M \to N$ (cross-bar) : $6\,\Omega$
• $N \to Q$ : $5\,\Omega$

The key observation: the apex $R$ is a dead-end junction with only two resistors attached, so current entering it from M must all leave to N. The path $M \to R \to N$ is therefore a plain series chain.

Step 2 — Collapse the series pair through the apex

Step 3 — Combine it with the cross-bar (parallel)

The 6 Ω chain $M{-}R{-}N$ and the 6 Ω cross-bar $M{-}N$ have the same two end nodes, M and N — that is precisely the definition of being in parallel:

(For two equal resistors in parallel the result is simply half of one of them: $6/2 = 3\,\Omega$.)

Step 4 — Add the two legs in series

Current entering at P has only one route: $P \to M$ (4 Ω), then through the M–N combination (3 Ω), then $N \to Q$ (5 Ω). Same current flows through all three in turn ⇒ series ⇒ resistances add:

Step 5 — Sanity-check the distractors

• (B) 21 Ω — what you get if you wrongly add everything in series ($4+3+6+3+5$), ignoring the parallel loop.
• (A) 1/12 Ω and (D) 1/21 Ω — these are the reciprocals of 12 and 21, i.e. the classic slip of forgetting to invert $1/R$ at the end of the parallel step. Note also that the answer must be larger than any single resistor here (the 4 Ω and 5 Ω legs alone already give 9 Ω), so any fraction-of-an-ohm answer is impossible on inspection.

ANSWER: C