
Figure shows three similar lamps ${L_1}$ ,\[{L_2}\] and ${L_3}$ connected across a power supply. If the lamp ${L_3}$ fuses, how will the light emitted by ${L_1}$ and ${L_2}$ change?
(A) No change
(B) Brilliance of ${L_1}$ decreases and that of ${L_2}$ increases
(C) Brilliance of both ${L_1}$ and ${L_2}$ increases
(D) Brilliance of both ${L_1}$ and ${L_2}$ decreases
Answer
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Hint: 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
${R_t} = R + R\parallel R$
Now we can write resistance connect in parallel as $\dfrac{{RR}}{{R + R}}$ hence the equation becomes
${R_t} = R + \dfrac{{RR}}{{R + R}}$
Now we can write the above equation as
${R_t} = R + \dfrac{R}{2}$
After simplifying the above equation we get
${R_t} = \dfrac{{3R}}{2}$
Now the current in the circuit is
$I = \dfrac{E}{{{R_t}}}$
Now substitute the value of ${R_t}$ in the above equation
$I = \dfrac{E}{{\dfrac{{3R}}{2}}}$
As we know that ${L_2}$ and ${L_3}$ are connected in parallel and have identical resistance so the total current $I$ is equally divided into the two branch. Thus, the current through ${L_2}$ or ${L_3}$ is $\dfrac{I}{2}$
When the lamp ${L_3}$ is fused, total resistance becomes $R{}_t' = R + R$
Then $R{}_t' = 2R$
Now the current ${I'} = \dfrac{E}{{2R}}$ Now this current passes through ${L_1}$ and ${L_2}$
Thus, the current in the ${L_1}$ decreases and current in the ${L_2}$ increases
Hence, the correct option is (B).
Brilliance of ${L_1}$ decreases and that of ${L_2}$ 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 $\Omega $
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
${R_t} = R + R\parallel R$
Now we can write resistance connect in parallel as $\dfrac{{RR}}{{R + R}}$ hence the equation becomes
${R_t} = R + \dfrac{{RR}}{{R + R}}$
Now we can write the above equation as
${R_t} = R + \dfrac{R}{2}$
After simplifying the above equation we get
${R_t} = \dfrac{{3R}}{2}$
Now the current in the circuit is
$I = \dfrac{E}{{{R_t}}}$
Now substitute the value of ${R_t}$ in the above equation
$I = \dfrac{E}{{\dfrac{{3R}}{2}}}$
As we know that ${L_2}$ and ${L_3}$ are connected in parallel and have identical resistance so the total current $I$ is equally divided into the two branch. Thus, the current through ${L_2}$ or ${L_3}$ is $\dfrac{I}{2}$
When the lamp ${L_3}$ is fused, total resistance becomes $R{}_t' = R + R$
Then $R{}_t' = 2R$
Now the current ${I'} = \dfrac{E}{{2R}}$ Now this current passes through ${L_1}$ and ${L_2}$
Thus, the current in the ${L_1}$ decreases and current in the ${L_2}$ increases
Hence, the correct option is (B).
Brilliance of ${L_1}$ decreases and that of ${L_2}$ 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 $\Omega $
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