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Question: Two lamps, each with a resistance of \[50\Omega \], are connected in series. The lamps will fuse if ...

Two lamps, each with a resistance of 50Ω50\Omega , are connected in series. The lamps will fuse if a power of more than 200  W200\;{\text{W}} is dissipated in it. Then the maximum voltage that can be applied to the circuit is:
A. 100V100V
B. 200V200V
C. 150V150V
D. 250V250V

Explanation

Solution

Since, the lamps are connected in a series combination, the resistances of both of these lamps will add up to form net equivalent resistance. This question can easily be solved by using the basic equation of power and voltage.

Formula used: We will use the following formula to find out our required solution:
Pmax=V2maxR{P_{\max }} = \dfrac{{{V^2}_{\max }}}{R}
Where,
Pmax{P_{\max }} is the maximum power
Vmax{V_{\max }} is the maximum voltage
RR is the net resistance

Complete step by step answer:
According to the question, it is provided to us that
Pmax{P_{\max }} before fusing =200W = 200W
Also, the lamps are connected in series. So, the net effective resistance will add up as they are in a series combination.
That is,
Req=R1+R2{\operatorname{R} _{eq}} = {R_1} + {R_2}
Where
Req{\operatorname{R} _{eq}} is the equivalent resistance of both the lamps
R1{R_1} is the resistance of the first lamp
R2{R_2} is the resistance of the second lamp
Therefore, we get
Req=50+50=100Ω{\operatorname{R} _{eq}} = 50 + 50 = 100\Omega
Now, we will use our main formula that is to be used to find out the maximum voltage that can be applied so that the lamps do not get fused.
That is,
Pmax=V2maxR{P_{\max }} = \dfrac{{{V^2}_{\max }}}{R}
On rearranging the above formula, we get
V2max=Pmax×R{V^2}_{\max } = {P_{\max }} \times R
Now, we will substitute the values of maximum power and the net equivalent resistance in the above formula to get
V2max=200×100=20000{V^2}_{\max } = 200 \times 100 = 20000
On further solving this equation, we get
Vmax=141.42V\therefore {V_{\max }} = 141.42V
Hence, the lamps will fuse if we apply a voltage greater than 141.42V141.42V. So, we can apply any amount of voltage below this value so that the lams do not fuse.
Looking at the options provided to us, the only option less than 141.42V141.42Vis option (A.) which is 100V100V.
So, the correct option is (A).

Note: The rate at which work is done or energy is transformed into an electrical circuit is electric power. Simply put, it is a measure of how much energy over a period of time is used.