Given a single-loop electric circuit with 2 light bulbs in series and a battery of course. I was wondering if it is possible for electrons travelling in the circuit to convert all their potential energy to light and heat for the 1 light bulb only, if not what dictates how much energy is converted at each light bulb? Does it have to do with the resistance of each light bulb? If so could you please explain further? And if all potential energy was converted to heat and light in one light bulb, would current still flow to the other terminal of the battery?
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1 Answer
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As a first approximation, you can treat light bulbs as Ohm's resistor. Current continuity, alias Kirchhoff current law, states that electric current is constant through the circuit.
Thus, voltage drop across each light bulb is $\Delta V_i = \dfrac{R_i}{\sum_k R_k} E$, while the current is $I = \dfrac{1}{\sum_k R_k} E$.
Power balance of each electric dipole, the generator and the bulbs gives us:
- generator: $P^{gen} = E I$
- light bulbs: $P^{abs}_i = I \Delta V_i = R_i I^2$,
so that, $I E = I (\Delta V_1 + \Delta V_2)$, and $P^{gen} = P^{abs}_1 + P^{abs}_2$. As you can see the power absorbed by each light bulb is proportional to its resistance, and
$P^{abs}_1 = \dfrac{R_1}{\sum_k R_k} P^{gen}$$\quad , \quad$$P^{abs}_2 = \dfrac{R_2}{\sum_k R_k} P^{gen}$.
