Explain the equivalent resistance of a series and parallel resistor network.

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Q. Explain the equivalent resistance of a series and parallel resistor network.

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Concept understanding — Resistors in Series and Parallel

Series combination. When resistors are connected end to end so the same current I is forced through every one of them in turn (since charge cannot accumulate anywhere along the chain), the voltage across each one differs according to Ohm's law ( $V_1=IR_1$ , etc.), and the supply voltage V is the SUM of these individual drops: $V=V_1+V_2+V_3=I(R_1+R_2+R_3)$ . This defines the equivalent series resistance,

R_S = R_1+R_2+R_3+\ldots

which is always GREATER than even the largest individual resistor in the chain.

Parallel combination. When resistors are instead all connected directly across the same two points, each one experiences the identical voltage V, but the total current I splits into separate branch currents ( $I_1=V/R_1$ , etc.) that must add back up to the total, $I=I_1+I_2+I_3$ . This gives the equivalent parallel resistance through the reciprocal-sum rule,

\dfrac{1}{R_P} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\ldots

which is always LESS than even the smallest individual resistor -- adding another parallel branch can only open up additional room for current to flow, never restrict it.

Why household wiring is parallel. Every appliance in a house is deliberately connected in parallel across the mains supply, precisely because a parallel connection guarantees each device gets the full supply voltage and operates completely independently of the others -- switching any one appliance off leaves the voltage across, and the current through, every other appliance completely unaffected. A series connection, by contrast, would mean switching off any single appliance breaks the circuit for every appliance downstream of it.

Reducing complex networks. However elaborate a resistor network looks -- a multi-section "ladder" of alternating series and parallel groups, or a symmetric bridge-like arrangement -- it can generally be solved by repeatedly identifying and collapsing obvious series or parallel sub-groups, working from the network's interior outward, until a single equivalent resistance remains. A special but very useful case is a network with a genuine geometric symmetry: if two points in the network are forced, by symmetry, to sit at exactly the same potential, then any resistor bridging those two points carries NO current at all and can simply be deleted from the network without changing its overall behaviour -- exactly the trick that also underlies the balance condition of the Wheatstone's bridge.

Step 1 (Series). When resistors $R_1, R_2, R_3$ are connected end to end, the same current I flows through all of them (charge cannot accumulate anywhere along the chain).

Step 2. By Ohm's law, the voltage drops across each differ: $V_1=IR_1$ , $V_2=IR_2$ , $V_3=IR_3$ ; the supply voltage V equals their sum, $V=V_1+V_2+V_3=I(R_1+R_2+R_3)$ .

Step 3. Writing $V=IR_S$ defines the series equivalent resistance, $R_S=R_1+R_2+R_3$ -- always GREATER than the largest individual resistor.

Step 4 (Parallel). When the same three resistors are instead all connected across the same two points, each sees the identical voltage V, but the total current I splits into branch currents $I_1=V/R_1$ , $I_2=V/R_2$ , $I_3=V/R_3$ , which sum to the total: $I=I_1+I_2+I_3=V\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\right)$ .

Step 5. Writing $I=V/R_P$ defines the parallel equivalent resistance through $\dfrac{1}{R_P}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$ -- always LESS than the smallest individual resistor.

Step 6. These two rules let any resistor network, however complex, be reduced by repeatedly collapsing obvious series or parallel sub-groups from the inside outward, until one single equivalent resistance remains for the whole network.

✓Final answer

R_S=R_1+R_2+R_3

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 55/6/11 markMCQ
$R_s$
›Reveal solutionSolution
For $n$ identical resistors $R$ , series equivalent $R_s = nR$ and parallel equivalent $R_p = R/n$ . Their difference is $R_s - R_p = \frac{n^2 - 1}{n}R$ , which matches option (C).

The problem is about combining identical resistors — a classic exercise in spotting how series and parallel formulas scale. When you put $n$ resistors of the same value $R$ in series, the total resistance just adds up: $R_s = nR$ . When you put them in parallel, the reciprocal adds up, giving $R_p = R/n$ . The question then asks for the difference $R_s - R_p$ , which is a simple subtraction — but the trick is in the algebra and in recognising which of the given options matches the result.

Let’s walk through it step by step.

Series combination: For $n$ resistors each of resistance $R$ connected end-to-end, the equivalent resistance is the sum:

$R_s = R + R + \cdots + R \quad (n \text{ times}) = nR.$

Parallel combination: For $n$ identical resistors connected across the same two points, the equivalent resistance is given by:

$\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \cdots + \frac{1}{R} \quad (n \text{ times}) = \frac{n}{R}.$
Taking the reciprocal:
$R_p = \frac{R}{n}.$

Find the difference: Subtract the parallel equivalent from the series equivalent:

$R_s - R_p = nR - \frac{R}{n}.$

Simplify the expression: Factor out $R$ and combine the terms over a common denominator:

$R_s - R_p = R\left(n - \frac{1}{n}\right) = R\left(\frac{n^2 - 1}{n}\right) = \frac{n^2 - 1}{n}R.$
Watch out

A common mistake is to forget that $R_p = R/n$ and instead write $R_p = nR$ (confusing parallel with series) or to mishandle the subtraction as $nR - R/n = \frac{n^2R - R}{n} = \frac{(n^2-1)R}{n}$ — which is correct, but then students sometimes mis-match it to an option like $\frac{n^2-1}{n^2}R$ by incorrectly dividing by an extra $n$ .

Tip

You can quickly check with a small number, say $n=2$ and $R=1\,\Omega$ . Then $R_s = 2\,\Omega$ , $R_p = 0.5\,\Omega$ , so $R_s - R_p = 1.5\,\Omega$ . Now test the options: (A) gives $3/4$ , (B) gives $5/3$ , (C) gives $3/2 = 1.5$ , (D) gives $5/4$ . Only (C) matches. This is a great sanity check in an exam.

✓Final answer
$(R_s - R_p)$
CBSE 2023Set ANNUAL1 markMCQ
$Q. There is a current of 1.0 A in the circuit shown below. What is the resistance of P ?$
(a) 3.5 Ω
(b) 1.5 Ω
(c) 4.5 Ω
(d) 2.5 Ω
›Reveal solutionSolution

Applying $R_{total}=V/I$ to the single series loop and subtracting the two known resistors gives $P = 4.5\ \Omega$ .

Working

The circuit is a single series loop: a 10 V battery, a $3\ \Omega$ resistor, a $2.5\ \Omega$ resistor, and the unknown resistor $P$ , carrying a current $I = 1.0$ A.

Since all elements are in series, applying Ohm's law to the whole loop (equivalent resistance = sum of series resistances):

$V = I(R_1+R_2+P)$

$10 = 1.0\times(3+2.5+P)$

$10 = 5.5 + P$

$P = 4.5\ \Omega$

✓Final answer
$P = 4.5\ \Omega$
CBSE 2020Set ANNUAL1 markMCQ
$Q. The current in the circuit is : [figure: a 15 V battery connected across three 15 Ω resistors, all three resistors in parallel with each other and with the battery]$
(a) 4 A
(b) 1 A
(c) 2 A
(d) 3 A
›Reveal solutionSolution

Three equal $15\ \Omega$ resistors in parallel combine to $5\ \Omega$ ; Ohm's law with the 15 V battery then gives a total current of 3 A.

Working

For three equal resistors $R=15\ \Omega$ in parallel:

$\dfrac1{R_{eq}}=\dfrac1{15}+\dfrac1{15}+\dfrac1{15}=\dfrac3{15}\ \Rightarrow\ R_{eq}=5\ \Omega$

Applying Ohm's law to the whole circuit (battery driving the parallel combination):

$I=\dfrac{V}{R_{eq}}=\dfrac{15}{5}=3\ \text{A}$

✓Final answer
$I=3$

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Q.Explain the equivalent resistance of a series and parallel resistor network.

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