Question

Two bulbs, A and B consume the same power when operated at voltage $V_A$ and $V_B$ respectively. Bulbs are connected with a supply of DC source, then

Column - I	Column -II
a. In series connection, the ratio of potential difference across A and B	$\frac{R_A}{R_B}$
b. In series connection the ratio of power consumed by A and B	$\frac{V_A^2}{V_B^2}$
c. In parallel connection, the ratio of current flowing through A and B	$\frac{R_B}{R_A}$
d. In parallel connection, the ratio of power connsumed in A and B	$\frac{V_B^2}{V_A^2}$

• A. a(i,ii), b(i,ii), c(iii, iv), d(iii, iv)
• B. a(i,ii), b(iii,iv), c(i, ii), d(iii, iv)
• C. a(iii,iv), b(iii,iv), c(i, ii), d(i, ii)
• D. a(iii, iv), b(i,ii), c(iii, iv), d(i,ii)

Answer 1

The correct option is A a(i,ii), b(i,ii), c(iii, iv), d(iii, iv)

The resistance of bulbs A and B are
$R_A=\frac{V_A^2}{P}$ and $R_B=\frac{V_B^2}{P}$
$\Rightarrow \frac{R_A}{R_B}=\frac{V_A^2}{V_B^2}$

If the bulbs are connected in series, the current through them will be the same. Hence, the potential difference across them will be proportional to their resistances. Therefore, the ratio of the potential difference across them will be equal to the ratio of their resistances $R_A:R_B$ which is $V_A^2:V_B^2$
So (a) $\to$(i,ii)\

The power consumed will be
$P_A=r^2{R}_A$ $P_B=r^2{R}_B$
$\frac{P_A}{P_B}=\frac{R_A}{R_B}=\frac{V_A^2}{V_B^2}$
Hence (b) $\to$(i,ii)

If bulbs are connected in parallel, then
$i_1{R}_A=i_2{R}_B$
$\frac{i_1}{i_2}=\frac{R_B}{R_A}=\frac{V_B^2}{V_A^2}$
So, (c) $\to$ (iii, iv)

and the power consumed
$\frac{P_A}{P_B}=\frac{V^2/R_A}{V^2/R_B}=\frac{R_B}{R_A}=\frac{V_B^2}{V_A^2}$
Hence (d)$\to$ (iii, iv)