Question
Question: Figure shows three similar lamps \({L_1}\) ,\[{L_2}\] and \({L_3}\) connected across a power supp...
Figure shows three similar lamps L1 ,L2 and L3 connected across a power supply. If the lamp L3 fuses, how will the light emitted by L1 and L2 change?
(A) No change
(B) Brilliance of L1 decreases and that of L2 increases
(C) Brilliance of both L1 and L2 increases
(D) Brilliance of both L1 and L2 decreases
Solution
Brightness or brilliance is an attribute of visual perception in which a source appears to be radiating or reflecting light. In other words, brilliance or brightness is the perception elicited by the luminance of a visual target. Brightness is also color coordinate in HSL color space: hue, saturation, and lightness, meaning here brightness
Complete step by step solution:
Let us consider R to be the resistance of each lamp. If emf E is the applied
Then the total resistance of the circuit is given by
Rt=R+R∥R
Now we can write resistance connect in parallel as R+RRR hence the equation becomes
Rt=R+R+RRR
Now we can write the above equation as
Rt=R+2R
After simplifying the above equation we get
Rt=23R
Now the current in the circuit is
I=RtE
Now substitute the value of Rt in the above equation
I=23RE
As we know that L2 and L3 are connected in parallel and have identical resistance so the total current I is equally divided into the two branch. Thus, the current through L2 or L3 is 2I
When the lamp L3 is fused, total resistance becomes Rt′=R+R
Then Rt′=2R
Now the current I′=2RE Now this current passes through L1 and L2
Thus, the current in the L1 decreases and current in the L2 increases
Hence, the correct option is (B).
Brilliance of L1 decreases and that of L2 increases.
Note: Resistance is defined as the opposition that a substance or material offers to the flow of electric current through it. It is denoted by the uppercase letter R. The standard unit of resistance is the ohm, sometimes written out as a word, and sometimes symbolized by the uppercase Greek letter omega Ω