Let's say that I have a circuit consisting of a power supply of $220V$ and $2$ bulbs of power-voltage rating $40W-220V$ and $60W-220V$ in series. How would I calculate the current flowing across the circuit? I know how to calculate it if there is a single bulb (Using $P = I^2 * R$, $P = V^2/R$ and $P = VI$), but I'm confused about the use of the formulae for multiple bulbs/appliances.
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$\begingroup$ I think there is information missing, what is the voltage rating of bulb ? @Twilight $\endgroup$– BrianCommented Mar 19, 2021 at 18:12
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$\begingroup$ Edited it now :) $\endgroup$– TwilightCommented Mar 19, 2021 at 18:22
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$\begingroup$ @ShreyanshPathak I see. But how do I calculate the resistance in this case? $\endgroup$– TwilightCommented Mar 19, 2021 at 18:22
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$\begingroup$ @ShreyanshPathak This is an answer. It's improper to post answers as comments. $\endgroup$– Bill NCommented Mar 19, 2021 at 19:44
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2 Answers
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An incandescent bulb is simply a resistor. The $P$-$V$ rating of a bulb is a statement about how much power the bulb would consume if placed in parallel with a $V$ voltage supply. Assuming the relationship of $$P=\frac{V^2}{R}$$ one can calculate the effective resistance of an individual bulb. The $P$ and $V$ used here are completely independent of any actual circuit, and are the numbers which tell you the resistance of the bulb.
In a diagram, put each bulb-resistor in its place and analyze the circuit to find actual individual currents, voltages, powers, etc.
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$\begingroup$ Thanks! So I need to used the rated voltage and power in the formula and not the actual voltage? And this formula is applicable to series circuits as well right? $\endgroup$– TwilightCommented Mar 20, 2021 at 4:23
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$\begingroup$ @Twilight yes the rated voltage would be used to calculate Resistance here as resistance would behave like the property of the material in this case as Temperature, length and Area are kept constant. $\endgroup$ Commented Mar 20, 2021 at 4:39
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The answer of Bill N is only a first approximation since in bulbs the resistance depends on temperature and so from Voltage. But if this is for school , the answer is the usual one.
