For all problems, assume that the battery voltage and lamp resistances are constant, no matter what current is present.
A 75-W lamp is connected to $125 V$.
- What is the current through the lamp?
A resistor is added to the lamp in the previous problem to reduce the current to half of its original value.
- What is the potential difference across the lamp?
I think that the potential difference across the lamp is just 125V. But in the solutions manual the answer is $6.3×10^1V$ $$V=I \cdot R=0.3 \cdot 2.1 \times 10^2=6.3 \times10^1 V$$.
I want to know why they use the resistance of the lamp before the change $(2.1×10^2)$ instead of the resistance after the change $(4.2×10^2)$?