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Let's say there's a parallel circuit consisting of some bulbs (which are lit up). If one of them fuses, I know that the others would continue to glow. But because one of the bulbs has fused, would the new resistance depend on the other bulbs? Would the current change accordingly (if so)?
Thank you
The new equivalent resistance would be change, not the individual resistance. Suppose there bulb in parallel with resistance $R_1$,$R_2$ and $R_3$. $$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$$
If one of them, says the third one, get fused, the equivalent resistance would change to $$\frac{1}{R'_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$$ So the resistance gets changed.
The current will change accordingly to resistance, the expression can be worked out.
That depends a lot on what type of bulb we are talking about (LED, incandescent, halogen, fluorescent, etc). For simplicity we (erroneously) assume that the bulb behaves like a linear resistor.
But because one of the bulbs has fused, would the new resistance depend on the other bulbs?
The new load resistance of all bulbs combined would be higher.
Would the current change accordingly (if so)?
The current delivered from the source would be lower.
The current through each remaining bulb will be slightly higher. This is caused by the internal impedance of the power source. Due to the lower source current, the voltage drop over the internal source impedance will also slightly lower, so the bulbs will see a slightly higher voltage. If the source impedance is very small as compared to the bulb resistance and/or the number of bulbs is large, this effect is negligible.
