Christmas lights

Well let's clear out the stuff that we already know. The total resistance for a number $N$ of resistors hooked up in parallel is given by...

$\frac{1}{\sum \Omega_{P}} = \frac{1}{\Omega_{1}} + \frac{1}{\Omega_{2}} + \frac{1}{\Omega_{3}} + ... \frac{1}{\Omega_{N}}$

The total resistance for a number $N$ of resistors hooked up is series is given by...

$\sum \Omega_{S} = \Omega_{1} + \Omega_{2} + \Omega_{3} + ... \Omega_{N}$

The power dissipated by a circuit is given by $P = I^{2}\sum\Omega$ , and we all know Ohm's law as $V = I\sum\Omega$ with $I$ being the current running through the circuit, $V$ being the voltage drop across the circuit, and $P$ being the power dissipated by the circuit.

The trick to solving this is understanding that the voltage drop across both of these circuits is the same , so what we can do is get both of their dissipated power functions/formulas in terms of $V$ , and then divide them both. I'll show you the math.

$\sum \Omega_{P} = \frac{1}{100} \Omega$

$\sum \Omega_{S} = 100 \Omega$

$P = \frac{V^{2}}{\sum \Omega}$

$\frac{P_{P}}{P_{S}} = \frac{\frac{V^{2}}{\sum \Omega_{P}}}{\frac{V^{2}}{\sum \Omega_{S}}} = \frac{\sum \Omega_{S}}{\sum \Omega_{P}} = \frac{100 \Omega}{\frac{1}{100} \Omega} = \boxed{10,000 \Omega}$

Analysing circuit P first, we see that the P.D. across each light is the same, and is equal to the voltage of the power supply. Hence, total power is given by

$100V^2/R$

In circuit S, the potential is equally divided across the 100 lights (as they have equal resistance); hence, total power is given by

$(100(V/100)^2)/R$

Hence,

$P_{P}/P_{S} = 100V^2/(V^2/100) = \boxed {10000}$ .

total resistance in circuit P is : $R_{P} = 1/100 Ω$

total resistance in circuit S is : $R_{S} = 100 Ω$

$P_{P}/P_{S} = (V^{2}/R_{P})/(V^{2}/R_{S}) = R_{S}/R_{P} = 100 / (1/100) = 10000$

Since $\text {Power} = \dfrac {V^2}{R} \propto \dfrac {1}{R}$ , we want the ratio of the net resistance in S to the net resistance in P. The net resistance in S is $1 \cdot 100 \, \Omega$ and the net resistance in P is $\dfrac {1}{100} \, \Omega$ , so our answer is $\boxed {10^4}$ .