Answer

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**Hint:**First we construct a circuit diagram of the given two bulbs connected in series. By using the equation power $(P) = \dfrac{{{V^2}}}{R}$ we will find the resistance of the respective bulb and by using that resistance we will obtain the Current $(I)$ flowing through the circuit. Now by using the ohm's law $V = IR$we will find the potential difference $(V)$ across each bulb.

Formula used:

$ \Rightarrow {P_A} = \dfrac{{{V^2}}}{{{R_A}}}$

$ \Rightarrow I = \dfrac{V}{R}$

**Complete step by step solution:**

Here we will first obtain the resistance of the respective bulb by using the formula of power $(P) = \dfrac{{{V^2}}}{R}$

For bulb A the resistance,

$ \Rightarrow {P_A} = \dfrac{{{V^2}}}{{{R_A}}}$

$ \Rightarrow {R_A} = \dfrac{{{V^2}}}{{{P_A}}}$ --------- Equation $(1)$

For Bulb a Power is $100W$ , Voltage is $200V$ , hence substituting the values of $P$ and $V$ in the equation $(1)$

$ \Rightarrow {R_A} = \dfrac{{{{(200)}^2}}}{{100}}$

$\therefore {R_A} = 400\Omega $

Similarly for Bulb B Power is 60 W, Voltage is 200 V, hence substituting the values of P and V in equation (1)

$ \Rightarrow {R_B} = \dfrac{{{{(200)}^2}}}{{60}}$

$\therefore {R_B} = 666.67\Omega $

Now as both the bulb are connected in series the total resistance ${R_{Total}}$of the circuit can be given as

$ \Rightarrow {R_{Total}} = {R_A} + {R_B}$

Putting the values in the above equation we get,

$ \Rightarrow {R_{Total}} = 400\Omega + 666.67\Omega $

$\therefore {R_{Total}} = 1066.67\Omega $

Now according to Ohm's law, the Voltage $(V)$ across any conductor is directly proportional to the current $(I)$ flowing through it at a constant temperature. Hence,

$ \Rightarrow V \propto I$

$ \Rightarrow V = RI$

Where R is constant of proportionality also known as resistance, Hence

$ \Rightarrow I = \dfrac{V}{R}$

$ \Rightarrow I = \dfrac{V}{{{R_{Total}}}}$ -----------equation $(2)$

Given $V = 200V$ across the circuit and ${R_{Total}} = 1066.67\Omega $ putting in the equation $(2)$

$ \Rightarrow I = \dfrac{{200V}}{{1066.67\Omega }}$

$ \Rightarrow I = 0.18799A \simeq 0.1875A$

Now the potential difference across each bulb A and B by using the formula $V = RI$

For bulb A the ${R_A} = 400\Omega $

$ \Rightarrow {V_A} = {R_A}I$

$ \Rightarrow {V_A} = 400 \times 0.1875 = 75\Omega $

For bulb B the ${R_B} = 667.67\Omega $

$ \Rightarrow {V_B} = {R_B}I$

$ \Rightarrow {V_B} = 666.67 \times 0.1875 = 125\Omega $

Here on comparing the potential difference of both bulb A and B

$\therefore {V_A} < {V_B}$

**Hence, option (C) is the correct answer.**

**Note:**Here we have to note that both the bulbs are connected in the circuits behaving as a source of resistance that’s why we used ohm's law. Similarly, if the bulbs are connected in parallel connection then we can find the total resistance ${R_{total}}$can be found by the formula$\dfrac{1}{{{R_{Total}}}} = \dfrac{1}{{{R_A}}} + \dfrac{1}{{{R_B}}}$.

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