(THE QUESTION IS FROM HEATING EFFECT OF CURRENT)

(MERITNATION EXPERT PLEASE ANSWER)

Two bulbs B1 and B2 are connected in series with an A.C source of emf 200V as shown.The labels on the bulbs reads 200V-60W and 200V-100W respectively .Calculate the ratio of the (a)the resistance of the bulb (R1/R2) (b) the power being consumed when connected in series (P1/P2) (c)the p.d across the bulbs (V1/V2)

(ans a,b,c =5:3)please make it simple,short,and undrstndn

A)

Power = V × I = V^{2}/R

Power of bulb B_{1} : Power of bulb B_{2} = 60 : 100

= > V^{2}/R_{1} : V^{2}/R_{2} = 3 : 5

= > R_{2} : R_{1} = 3 : 5

= > R_{1} : R_{2} = 5 : 3

B)

When connected in series the current through the two bulbs will be same. And the sum of voltages across the two bulbs is equal to 200 V.

Now,

P_{1} : P_{2} = I^{2}R_{1} : I^{2}R_{2}

= > P_{1} : P_{2} = R_{1} : R_{2} = 5 : 3

C)

Since the bulbs are connected in series, to find the ratio of PD across the bulbs we shall use the following relation,

P = V^{2}/R

Now,

P_{1} : P_{2} = (V_{1})^{2}/R_{1} : (V_{2})^{2}/R_{2}

= > 5/3 = (V_{1})^{2}/(V_{2})^{2} × R_{2}/R_{1}

= > (V_{1})^{2}/(V_{2})^{2} = (5/3)^{2}

= > V_{1} : V_{2} = 5 : 3

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