Two resistances, a 60 ohm and an unknown one are connected to a power source of $$120\ \text{V}$$ in a series arrangement. This way the power of the unknown resistance is $$60\ \text{W}$$. What is the Unknown resistance?

Correct Answer: 60

Solution

Let the unknown resistance be $$R$$. The current is $$I = \frac{120}{60 + R}$$, so the power in $$R$$ is $$\frac{14400R}{(60 + R)^2} = 60$$. This gives $$(60 + R)^2 = 240R$$, i.e. $$R^2 - 120R + 3600 = 0$$, or $$(R - 60)^2 = 0$$. Hence $$R = 60\ \Omega$$.

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