Question

Given figure shows arrangement of six bulbs and two battries each bulb is rated at 100V and there power is shown in figure
List I gives bulbs and list II gives there output power

With $S_2$ closed and $S_1$ open.
$\begin{array}{cc}\text{List~I}& \text{List~II}\\ \text{I)}{B}_1& \text{P)}\frac{256}{9}\\ \text{II)}{B}_2& \text{Q)}\frac{128}{9}\\ \text{III)}{B}_3& \text{R)}\frac{200}{9}\\ \text{IV)}{B}_4& \text{S)}\frac{512}{9}\\ & \text{T)}\frac{800}{9}\\ & \text{U)}\frac{400}{9}\end{array}$

• A. $I\to P,II-R,III\to S,IV-Q$
• B. $I\to P,II\to S,III\to P,IV\to Q$
• C. $I-Q,II-T,III-P,IV-U$
• D. $I-Q,II\to T,III\to U,IV\to R$

Answer 1

The correct option is B $I\to P,II\to S,III\to P,IV\to Q$

$I\to P,II-SIII-P,IV-Q$
Since $P=\frac{V^2}{R}\Rightarrow R=\frac{V^2}{P}$
Resistance of Bulbs $B_1,B_2,B_3\&B_4$ are $R_1=\frac{{100}^2}{100}=100\Omega ,$ $R_2=\frac{{100}^2}{50}=200\Omega$ , $R_{100w}=100\Omega$
$R_3=\frac{{100}^2}{25}=400$, $R_4=\frac{{100}^2}{50}=200\Omega$, $R_{200w}=50\Omega$

with $S_2$ closed and $S_1$ open
$\epsilon =i{R}_{eq}\Rightarrow 200=i\times 250\Rightarrow i=\frac{4}{5}A$
$P_1=(\frac{8}{15}{)}^2\times 100=\frac{256}{9}{P}_2=(\frac{8}{15}{)}^2\times 200=\frac{512}{9}$
$P_3=\left(\frac{4}{15}\right)\times 400=\frac{256}{9}{P}_4=(\frac{4}{15}{)}^2\times 200=\frac{128}{9}$