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 switch $S_1$ closed and switch $S_2$ 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 - R, II - U, III - T, IV - R
• B. I - Q, II - S, III - U, IV - T
• C. I - U, II - T, III - U, IV - R
• D. I - R, II - U, III - T, IV - P

Answer 1

The correct option is C I - U, II - T, III - U, IV - R

$I-U,II-T,III-U,IV-R$
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 $S_1$ closed $S_2$ open $\Rightarrow \epsilon =i{R}_{eq}\Rightarrow 300=i\times 300$
$i=1A\Rightarrow$ Power output of bulbs is $P=i^{2}R$
$P_1=(\frac{2}{3}{)}^2\times 100=\frac{400}{9}w{P}_2=(\frac{2}{3}{)}^2\times 200=\frac{800}{9}w$
$P_3=(\frac{1}{3}{)}^2\times 400=\frac{400}{9}w{P}_4=(\frac{1}{3}{)}^2\times 200=\frac{200}{9}w$