For the circuit shown, a shorting wire of negligible resistance is added to the circuit between points, A and B. when this shorting wire is added, bulb $$3$$ goes out. which $$bulb(s)$$ in the circuit brighten? All bulbs are identical.

Question

For the circuit shown, a shorting wire of negligible  resistance is added to the circuit between points, A and B. when this shorting wire is added, bulb $$3$$ goes out. which $$bulb(s)$$ in the circuit brighten? All bulbs are identical.

Infinity Learn · Accepted Answer

Let the resistance of each bulb be R. Initially: Req $$=$$ $$5R/3$$,    Finally : Req $$=$$ $$3R/2$$.Equivalent resistance decreases, so current increases in circuit and in '$$1$$' also. Hence brightness of '$$1$$' increases. It means potential difference across '$$1$$' increases, so across '$$2$$' potential difference decreases, hence brightness of $$2$$ decreases. Initially: Potential difference across '$$4$$'$$V4i$$ $$=$$  $$12$$  $$2R/3$$ε$$2R/3+R=$$ε$$5Finally$$:  $$V4f$$  $$=$$ $$R/2$$ε$$R/2+R$$ $$=$$ε$$3Since$$ $$V4f$$ > $$V4i$$, hence brightness of $$4$$ increases.

For the circuit shown, a shorting wire of negligible resistance is added to the circuit between points, A and B. when this shorting wire is added, bulb 3 goes out. which bulb(s) in the circuit brighten? All bulbs are identical.

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