A rating is a resistance in disguise
Every rating problem starts the same way: turn the rating into a resistance. A bulb rated \(P\) at \(V\) has \(R = V^2/P\). Bulb A, rated \(100\,\text{V}\), \(100\,\text{W}\), has \(100^2/100 = 100\,\Omega\). Bulb B, rated \(100\,\text{V}\), \(50\,\text{W}\), has \(100^2/50 = 200\,\Omega\).
For two bulbs rated at the same voltage, \(R \propto 1/P\). The lower-rated bulb has the larger resistance, because it must take less current at the same voltage. A \(60\,\text{W}\) and a \(100\,\text{W}\) bulb, both for \(200\,\text{V}\), have \(667\,\Omega\) and \(400\,\Omega\), in the ratio \(5 : 3\).
A \(25\,\text{W}\) and a \(100\,\text{W}\) bulb are both rated for \(220\,\text{V}\), with filaments of the same length. Which has the thicker filament?
Bulbs in series
Two bulbs are in series when they sit one after the other on a single path. Every charge that passes one must pass the other, so the same current \(I\) flows through both. The supply voltage splits between them: \(V = IR_1 + IR_2\). So the pair passes the current of a single resistance \(R_1 + R_2\). The next chapter treats such combinations fully.
Bulb A is rated \(100\,\text{V}\), \(100\,\text{W}\) and bulb B \(100\,\text{V}\), \(50\,\text{W}\). They are joined in series across \(150\,\text{V}\). Which glows brighter?
In series across \(150\,\text{V}\), the pair has \(100 + 200 = 300\,\Omega\) and carries \(0.5\,\text{A}\). Bulb A gets \(25\,\text{W}\), a quarter of its rating, and glows dimly. Bulb B gets \(50\,\text{W}\), exactly its rating, and glows normally. It also has \(0.5 \times 200 = 100\,\text{V}\) across it, its rated voltage.
Bulbs of \(60\,\text{W}\) and \(100\,\text{W}\), both rated for \(200\,\text{V}\), are joined in series across \(200\,\text{V}\). Find the ratio of their powers, which glows brighter, and the total power.
\(R_{60} = 200^2/60 = 667\,\Omega\) and \(R_{100} = 200^2/100 = 400\,\Omega\). The same current flows through both.
\(P = I^2R\) with one \(I\), so \(P_{60} : P_{100} = R_{60} : R_{100}\). The current is \(I = 200/(667 + 400)\).
\(P_{60} : P_{100} = 667 : 400 = 5 : 3\). \(I = 200/1067 = 0.1875\,\text{A}\), so the total is \(200 \times 0.1875 = 37.5\,\text{W}\).
The powers are in the ratio \(5 : 3\), so the \(60\,\text{W}\) bulb glows brighter. Together they give only \(37.5\,\text{W}\).
\[ R_1 = \dfrac{V^2}{P_1}, \qquad R_2 = \dfrac{V^2}{P_2} \]
Why. both bulbs are rated for the voltage \(V\).
\[ P = \dfrac{V^2}{R_1 + R_2} \]
Why. in series across \(V\) the pair acts as one resistance \(R_1 + R_2\).
\[ P = \dfrac{1}{1/P_1 + 1/P_2} = \dfrac{P_1 P_2}{P_1 + P_2} \]
Read it. put in step 1 and cancel \(V^2\).
For \(40\,\text{W}\) and \(60\,\text{W}\) bulbs this gives \(40 \times 60/100 = 24\,\text{W}\) in all. For two equal \(100\,\text{W}\) bulbs it gives \(50\,\text{W}\), \(25\,\text{W}\) each: each has half its voltage, so a quarter of its power. Two bulbs in series always give less than either would alone.
Two bulbs rated \(220\,\text{V}\), \(100\,\text{W}\) and \(220\,\text{V}\), \(25\,\text{W}\) are joined in series across \(220\,\text{V}\). What is the total power, in watts?
Bulbs in parallel
Two bulbs are in parallel when each is joined directly across the same two points. Each then has the full supply voltage across it. If that is the rated voltage, each bulb gives exactly its rated power, whatever the other does. The currents from the supply add.
Take the chip "parallel" in the figure above and set \(100\,\text{V}\). A gives \(100\,\text{W}\) and B gives \(50\,\text{W}\), \(150\,\text{W}\) in all. The currents are \(1\,\text{A}\) and \(0.5\,\text{A}\), so the supply gives \(1.5\,\text{A}\). Now the higher-rated bulb glows brighter, as its label suggests. Three \(100\,\text{W}\) bulbs in parallel at their rated voltage give \(300\,\text{W}\). Raise the supply above \(100\,\text{V}\) and both bulbs are driven beyond their ratings.
That is why the wiring of a house is parallel. Every lamp, fan and heater sits across the full \(220\,\text{V}\) and gives its rated power. Each can be switched on or off alone. A failed bulb breaks only its own branch; in series it would cut every device.
In Fig 6.1 all three resistors are \(4\,\Omega\). Find the current from the battery, the power in each resistor and the total power. Which resistor heats up fastest?
\(R_2\) and \(R_3\) are in parallel: they have the same voltage and are equal, so they share the current equally. The pair passes twice the current of one at the same voltage, so it acts as \(2\,\Omega\).
The battery sees \(R_1\) in series with the pair: \(4 + 2 = 6\,\Omega\), and \(I = 12/6\).
\(I = 2\,\text{A}\). \(P_1 = 2^2 \times 4 = 16\,\text{W}\). Each of \(R_2\) and \(R_3\) carries \(1\,\text{A}\), so \(P_2 = P_3 = 1^2 \times 4 = 4\,\text{W}\).
The battery gives \(2\,\text{A}\) and \(24\,\text{W}\) in all. \(R_1\) carries the whole current and heats fastest, four times as fast as each of the others.
Three bulbs of \(100\,\text{W}\), \(60\,\text{W}\) and \(40\,\text{W}\), each rated for \(200\,\text{V}\), are joined in parallel across \(200\,\text{V}\). What current does the supply give, in amperes?
Kettle coils, fuses and safe ratings
A kettle has two equal coils, and either alone boils its water in \(10\) minutes. The heat needed is fixed, so the time goes as \(1/P = R/V^2\) at the mains voltage. In series the coils make \(2R\): half the power, so \(20\) minutes. In parallel they make \(R/2\): twice the power, so \(5\) minutes. Cut one coil in half and use half: \(R/2\) again, twice the power, half the time.
A fuse is a short wire that melts when its current passes a set value, and breaks the circuit. It sits in the main wire, so it carries the sum of all the branch currents. A \(1000\,\text{W}\) heater on \(200\,\text{V}\) draws \(5\,\text{A}\), and a \(100\,\text{W}\) bulb \(0.5\,\text{A}\).
With the heater alone the fuse carries exactly \(5\,\text{A}\) and holds. Switch the bulb on as well and it carries \(5.5\,\text{A}\) and melts. A fuse of the right rating protects the wiring from overheating; a thicker wire put in its place protects nothing.
A resistor carries a power rating as well as its resistance. A \(100\,\Omega\) resistor rated \(1\,\text{W}\) may take at most \(I = \sqrt{P/R} = 0.1\,\text{A}\) and \(V = \sqrt{PR} = 10\,\text{V}\). Two in series share one current, so the pair can take \(20\,\text{V}\). Two in parallel share one voltage, so they can take only \(10\,\text{V}\), and then give \(2\,\text{W}\) in all.
One coil of a heater boils some water in \(12\) minutes; a second coil boils it in \(24\) minutes. How long do the two take when joined in parallel across the same supply?
Two wires of the same metal and length have radii \(r\) and \(2r\). Which statements are true?
Older Indian homes had porcelain fuse holders with a strand of fuse wire, which had to be replaced by hand after it melted. Most homes now have miniature circuit breakers in the meter box. An MCB trips a switch when its current stays above its rating, and it can be reset with a flick. Lighting circuits commonly use a small rating, such as \(6\,\text{A}\). The circuit of a geyser or an air conditioner uses a larger one, such as \(16\,\text{A}\).
- Expecting the higher-rated bulb to be brighter in series.In series one current flows, \(P = I^2R\), and the larger resistance, the lower rating, glows brighter.
- Using the rated power at a voltage other than the rated one.Turn the rating into \(R = V^2/P\), then find the power at the actual voltage.
- Adding the safe voltages of resistors joined in parallel.In parallel they share one voltage, so the pair is limited by the smaller safe voltage.
- Adding the times of two coils joined in parallel.In parallel the powers add, so the rates of heating add and the time falls.