B_{1}, B_{2} and B_{3} are three identical bu | vedclass

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(C) Let the resistance of each identical bulb be $R$ and the $EMF$ of the battery be $E$ with internal resistance $r$.When key $K$ is closed, bulbs $B

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Brightness of the bulb $B_{1}$ decreases and $B_{2}$ increases

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(C) Let the resistance of each identical bulb be $R$ and the $EMF$ of the battery be $E$ with internal resistance $r$.When key $K$ is closed, bulbs $B_{2}$ and $B_{3}$ are in parallel, and this combination is in series with $B_{1}$. The equivalent resistance is $R_{eq} = R + (R/2) = 1.5R$. The total current is $I = E / (1.5R + r)$. The voltage across $B_{1}$ is $V_{1} = I \cdot R = E \cdot R / (1.5R + r)$.When key $K$ is opened, bulb $B_{3}$ is disconnected. The circuit now consists of $B_{1}$ and $B_{2}$ in series. The new equivalent resistance is $R'_{eq} = 2R$. The new total current is $I' = E / (2R + r)$.Comparing the currents: Since $2R > 1.5R$, the total current $I'$ is less than $I$. Thus, the current through $B_{1}$ decreases, so its brightness decreases.The voltage across $B_{2}$ is $V'_{2} = I' \cdot R = E \cdot R / (2R + r)$. Comparing $V'_{2}$ with the previous voltage across $B_{2}$ (which was $V_{2} = I \cdot (R/2) = E \cdot (R/2) / (1.5R + r) = E \cdot R / (3R + 2r)$), we find that $V'_{2} > V_{2}$. Therefore, the brightness of $B_{2}$ increases.

$B_{1}$, $B_{2}$ and $B_{3}$ are three identical bulbs connected to a battery of steady $EMF$ with key $K$ closed. What happens to the brightness of the bulbs $B_{1}$ and $B_{2}$ when the key $K$ is opened?

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