Electric circuits
Flip a switch and a bulb lights instantly — but nothing about that "instant" glow is magic. It works only because charge has an unbroken loop to flow around: from the cell, through the wire, through the key, through the bulb, and back to the cell. Break that loop anywhere — pull a wire, open the key — and the glow dies at once, even though every component is still perfectly intact. This article builds the vocabulary and the one equation that describe every simple circuit you will meet in Class 9: the standard symbols used to draw cells, bulbs, keys, resistors, rheostats, ammeters and galvanometers; the difference between an open circuit and a closed circuit; and Ohm's Law, , applied to a single resistor. It also shows how a rheostat lets you dim or brighten a bulb by sliding a contact along a wire — nothing more exotic than changing resistance. Series and parallel combinations, power, and household wiring are Class X topics and are deliberately left out here.
A simple electric circuit is a closed conducting loop — cell, key, wires and a load — in which current flows only when the path is unbroken, and Ohm's Law links the potential difference, current and resistance of any single resistor in that loop.
The core idea
What makes a circuit a circuit. At minimum you need a source of electrical energy (a cell), a device that uses it (a bulb, say), and connecting wires joining them into one continuous loop. A key (switch) is added purely for convenience, so the loop can be broken and remade without disconnecting wires by hand.
Open versus closed. The circuit is closed when the conducting path runs unbroken all the way around the loop — charge can flow continuously, and the bulb glows steadily. The circuit is open the moment that path is interrupted anywhere: the key is switched off, a wire slips out of a terminal, or a component fails. An open circuit carries zero current everywhere in the loop, not just at the break — because the whole loop is one connected path, a single gap stops the flow throughout it.
Reading a circuit diagram. Real components are drawn as standard symbols rather than sketched realistically, so anyone can read a diagram at a glance: a cell is one long thin line (positive) next to one short thick line (negative); a bulb is a circle with a cross or filament inside; a key/switch is a gap in the wire with a small circle at each end, bridged by a short diagonal line when closed; a resistor (fixed resistance wire, or a resistance box) is a rectangle or zigzag; a rheostat is a resistor symbol with an arrow drawn across it, showing its resistance can be varied; an ammeter is a circle marked A, always wired so the current being measured passes straight through it; a galvanometer is a circle marked G, used to detect small currents rather than measure large ones precisely.
Ohm's Law for a single resistor. Within a simple loop containing just one resistor, potential difference, current and resistance are tied together by
where: is the potential difference across the resistor (volts), is the current flowing through it (amperes), and is its resistance (ohms). Given any two of these, the third follows directly — this is the one calculation this topic asks of you, applied to a single conductor between two points, not to a network of several resistors.
Why a rheostat is useful. A rheostat is resistance wire with a sliding contact; moving the contact changes how much of that wire lies in the circuit, which changes . By Ohm's Law, changing while the cell's voltage stays fixed changes — which is exactly how a bulb's brightness is turned up or down in a simple circuit, without touching the cell itself. Other DC sources, such as accumulators (rechargeable cells), appear in this syllabus only by name; their internal working is not required.
The goal of this derivation is to show, starting only from the statement and one set of measured circuit values, exactly how to isolate whichever quantity — the potential difference, the current, or the resistance — is the unknown in a given problem, and to make explicit why the other two quantities must be measured (or given) before the third can ever be calculated for a simple loop holding a single resistor between two points.
Step 1. Start from Ohm's Law as it applies to one resistor in a closed loop: . This says the potential difference across the resistor equals the current through it multiplied by its resistance — nothing here requires knowing how many other components share the loop, only that a single defined resistor sits between the two points where is measured.
Step 2. To find the current when and are known, divide both sides by : . To find the resistance when and are known, divide both sides by instead: . Each rearrangement is the same equation, read for a different unknown — there is no separate formula to memorise beyond itself.
Step 3. Apply the correct rearrangement to the numbers given. A key/switch and connecting wires are assumed to have negligible resistance of their own, so all of the loop's resistance is credited to the named resistor (or bulb) — this is why the circuit is treated as "a single resistor" even though a real bulb and a real key are also present.
Step 4. Check the answer's size against common sense: household cells and simple lab circuits typically deliver a few volts and currents of a few tenths to a few amperes, so a computed voltage of, say, or a current of is physically reasonable, while an answer of from the same small cell would signal an arithmetic slip, not a real result.
Worked example
Question: A circuit consists of a cell, a key, a resistor of resistance 8 Ω, and a bulb (its resistance is included in the 8 Ω figure), all in a single loop. With the key closed, an ammeter in the circuit reads a current of 0.5 A. Calculate the potential difference across the resistor.
Step 1. Write Ohm's Law for this single resistor: .
Step 2. Substitute the given values, and :
Answer: the potential difference across the resistor is 4 V.
Competitive stretch
Suppose the same cell is now connected across a rheostat instead of the fixed 8 Ω resistor, and the student slides the contact so that only 5 Ω of resistance wire remains in the circuit, while the cell's potential difference stays at the 4 V found above. Using , the new current is — larger than the original 0.5 A, because reducing the resistance with the rheostat's slider let more current flow for the same potential difference, and the bulb would glow brighter as a result.
Common traps
- Drawing a bulb and forgetting the key, or wiring the ammeter so current does not pass straight through it — an ammeter must sit directly in the current's path, not off to the side.
- Mixing up a rheostat with a cell: a rheostat only changes resistance (and hence current); it stores no charge and supplies no energy of its own.
- Applying to several resistors in series or parallel without first finding their equivalent resistance — that combination step is Class X; here, applies to one resistor at a time.
- Reading "closed" circuit as the everyday sense of "shut off." In circuit language, closed means the path is complete and current can flow — the opposite of a closed door blocking passage.
Ask the tutor: Want to practice drawing a specific circuit diagram or work through a few more Ohm's Law problems together?
Why does opening the key stop current everywhere, not just near the key?
A simple circuit is one single connected loop, so there is only one path charge can take all the way around. Opening the key removes the only path, and since the same charge must move together around the entire loop for current to exist anywhere in it, breaking the loop at any single point — near the key or far from it — drops the current to zero at every point simultaneously, not just locally.
Is a rheostat the same thing as a resistor?
Both are made of resistance wire and both resist current, but a resistor (or resistance box) has a fixed value you cannot change while it sits in the circuit, whereas a rheostat has a sliding contact that lets you dial its resistance up or down while the circuit is running. That is why a rheostat, not a fixed resistor, is the component used to dim or brighten a bulb smoothly.
Can I use V = IR on a circuit that has a resistor and a bulb both in the same loop?
At this level, yes, as long as you are treating them as combining into one effective resistance value that already accounts for both — exactly as the worked example does by saying the bulb's resistance is included in the 8 Ω figure. Working out the equivalent resistance of separate named resistors combined in series or parallel is a Class X skill; Class 9 keeps you to a single resistor value standing in for the whole loop.