Q.Suppose there is a circuit consisting of only resistances and batteries and we have to double (or increase it to $n$-times) all voltages and all resistances. Show that currents are unaltered. Do this for circuit of Example 3.7 in the NCERT Text Book for Class XII.

Concept: Linearity of Kirchhoff's laws under uniform scaling

Kirchhoff's voltage law in any loop reads $\sum \varepsilon = \sum IR$. If every emf $\varepsilon \to n\varepsilon$ and every resistance $R \to nR$, this becomes $n\sum\varepsilon = \sum I'(nR) = n\sum I' R$; dividing by $n$ recovers the original equation, so the same currents $I'=I$ still solve it. The junction rule involves only currents, so it is untouched. Hence scaling all emfs and all resistances by the same factor $n$ leaves every branch current unchanged. …

The general proof

A network of resistors and batteries is solved by two rules:

• Junction rule (KCL): at every node, $\sum I = 0$ — this equation involves only currents, no emfs or resistances.
• Loop rule (KVL): around every closed loop, $\sum \varepsilon = \sum IR$.

Now replace every emf $\varepsilon \to n\varepsilon$ and every resistance $R \to nR$ (same $n$ everywhere), and let $I'$ denote the new currents. Substituting into a loop equation:

Dividing throughout by $n$:

which is exactly the original loop equation with $I'$ in place of $I$. Since the junction equations are unaffected by the scaling (they contain no $\varepsilon$ or $R$ at all), the scaled network obeys the identical set of equations as the original — and because a well-posed resistor network has a unique current solution, $I' = I$ for every branch. Physically: each current is a ratio $I=(\text{driving emf})/(\text{resistance})$, and stretching numerator and denominator by the same factor leaves the ratio unchanged.

Applying it to Example 3.7's actual network

Example 3.7 (Fig. 3.17) is the bridge network $ABCD$ with:

• arms $AB = AD = 4\,\Omega$
• arms $BC = CD = 2\,\Omega$
• diagonal $AC$: a $1\,\Omega$ resistor in series with a $10\,\text{V}$ cell
• diagonal $BD$: an (ideal) $5\,\text{V}$ cell

Solving this network by Kirchhoff's rules (junction + loop equations, or equivalently node potentials with $V_D=0$) gives $V_A=7.5\,\text{V}$, $V_B=5\,\text{V}$, $V_C=0\,\text{V}$, and hence the branch currents

$I_{AC}=2.5\,\text{A},\quad I_{AB}=0.625\,\text{A},\quad I_{AD}=1.875\,\text{A},\quad I_{BC}=2.5\,\text{A},\quad I_{CD}=0\,\text{A},\quad I_{BD}=1.875\,\text{A}.$ …

Method: Proving a Circuit Property Holds for ANY Network, Then Verifying on One Example

Use this two-part method whenever a problem asks you to (a) prove a general property of resistor-battery circuits algebraically, AND (b) confirm it on one specific worked example from the textbook — the general proof and the numerical check are different skills, and both are required.

Steps

Step 1: Identify which of Kirchhoff's two rules involves the quantity you're scaling

The junction rule ($\sum I = 0$ at a node) involves only currents — no emfs or resistances — so it is automatically unaffected by scaling them. The loop rule ($\sum \varepsilon = \sum IR$) involves both, so that is the equation to examine.

Step 2: Substitute the scaled quantities into the loop rule and factor

Replace every emf $\varepsilon \to n\varepsilon$ and every resistance $R \to nR$ (same $n$ throughout), with $I'$ the new currents:

Dividing both sides by $n$ recovers exactly the original loop equation, with $I'$ in place of $I$.

Step 3: Invoke uniqueness to conclude the currents are unchanged …