Question

plzzz answer with explanation...
Don't only answer the option...
give the explanation too...​

Answer 1

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❏ Question:-

@ Three identical bulbs are connected in as shows in the figure . When switch S is closed the power consumed by the bulb B is P . What will be the power consumed by the same bulb when the switch S is opened ?

Diagram Refer to the questionar attachment ,

❏ Solution:-

✏ Given:-

The bulbs are identical So, the resistance of each bulb is same ,

let , the resistances of the bulbs A , B , C are

$\sf{R_A}$ = $\sf{R_B}$ = $\sf{R_C}$ = R (let )

When switch S is closed the power consumed by the bulb B is P .

✏ To Find:-

Power consumed by the bulb B when the switch S is opened .

✏ Explanation :-

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➩ $\underline{\boxed{\text{When S is closed}}}$

In this case , current passes through the all bulbs ( A , B and C ) .

Now , Effective Resistance of the whole circuit is

$\implies R'_{\red{closed}}= R_A+ \dfrac{R_B\times R_C}{R_B+R_C}$

$\implies R'_{\red{closed}}= R+ \dfrac{R\times R}{R+R}$

$\implies R'_{\red{closed}}= R+ \dfrac{ R^2}{2R}$

$\implies R'_{\red{closed}}= R+ \dfrac{ R}{2}$

$\implies R'_{\red{closed}}= \dfrac{2R+ R}{2}$

$\implies R'_{\red{closed}}= \dfrac{3R}{2}$

∴ Current Through the circuit ,

$\implies I_{\red{closed}}= \dfrac{E}{R'_{\red{closed}}}$

$\implies I_{\red{closed}}= \dfrac{E}{\dfrac{3R}{2}}$

$\implies I_{\red{closed}}= \dfrac{2E}{3R}$

∴ Current Through the Bulb B,

$\implies I_B=I_{\red{closed}}\times\dfrac{R_C}{R_B+R_C}$

$\implies I_B= \dfrac{2E}{3R}\times\dfrac{R}{R+R}$

$\implies I_B= \dfrac{\cancel{2}E}{3\cancel{R}}\times\dfrac{\cancel{R}}{\cancel{2}R}$

$\implies I_B= \dfrac{E}{3R}$

Hence ,

Power consumed in the bulb B is

$\implies( P_B)_{\red{closed}}= I_B^2R_B$

$\implies( P_B)_{\red{closed}}=\Big(\dfrac{E}{3R}\Big)^2\times R$

$\implies( P_B)_{\red{closed}}=\dfrac{E^2}{9\cancel{R^2}}\times\cancel{ R}$

$\implies( P_B)_{\red{closed}}=\dfrac{E^2}{9R}$

Now according to the question ,

$\implies( P_B)_{\red{closed}}= P$

$\implies\dfrac{E^2}{9R} = P$

$\implies\boxed{ E^2= 9PR}$ ---------(1)

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$\underline{\boxed{\text{When S is free}}}$

In this case Bulb B is short circuited , So current only passes through the Bulbs A and B , But no current pass through the Bulb C ,

Now , Effective Resistance of the whole circuit is

$\implies R'_{\red{free}}= R_A+R_B$

$\implies R'_{\red{free}}= R+R$

$\implies R'_{\red{free}}=2R$

∴ Current Through the circuit ,

$\implies I_{\red{free}}= \dfrac{E}{R'_{\red{free}}}$

$\implies I_{\red{free}}= \dfrac{E}{2R}$

$\implies I_{\red{free}}= \dfrac{E}{2R}$

∴ Current Through the Bulb B

$\implies I_B= I_{\red{free}}=\dfrac{E}{2R}$

Hence ,

Power consumed in the bulb B is

$\implies( P_B)_{\red{free}}= I_B^2R_B$

$\implies( P_B)_{\red{free}}=\Big(\dfrac{E}{2R}\Big)^2\times R$

$\implies( P_B)_{\red{free}}=\dfrac{E^2}{4\cancel{R^2}}\times\cancel{ R}$

$\implies( P_B)_{\red{free}}=\dfrac{E^2}{4R}$

$\implies( P_B)_{\red{free}}=\dfrac{9P\cancel{R}}{4\cancel{R}}$

$\implies\large{\boxed{( P_B)_{\red{free}}=\dfrac{9P}{4}}}$

➩ Option (1) $\implies\large{\boxed{( P_B)_{\red{free}}=\dfrac{9P}{4}}}$

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