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Physics Question on Current electricity

A 100 W bulb B1B_1 and two 60 W bulbs B2 and B3B_2 \, and\, B_3, are connected to a 250 V source, as shown in the figure. Now W1,W2 and W3W_1, W_2 \, and \, W_3 are the output powers of the bulbs B1,B2B_1, B_2 and B3B_3 respectively. Then,

A

W1>W2=W3W_1 > W_2 = W_3

B

W1>W2>W3W_1 > W_2 > W_3

C

W1<W2=W3W_1 < W_2 = W_3

D

W1<W2<W3W_1 < W_2 < W_3

Answer

W1<W2<W3W_1 < W_2 < W_3

Explanation

Solution

P=V2R  so,      R=V2PP = \frac{V^2}{R} \ \\\ so, \ \ \ \ \ \ R = \frac{V^2}{P}
∴           V2R   and R2=R3=V260\therefore \ \ \ \ \ \ \ \ \ \ \ \frac{V^2}{R} \ \ \ and \ R_2 = R_3 = \frac{V^2}{60}
Now,         W1=(250)2(R1+R2)2.R1Now, \ \ \ \ \ \ \ \ \ W_1 =\frac{(250)^2}{(R_1 + R_2)^2} . R_1
            W2=(250)2(R1+R2)2.R2 and W3=(250)2R3\ \ \ \ \ \ \ \ \ \ \ \ W_2 =\frac{(250)^2}{(R_1 + R_2)^2} . R_2 \ and \ W_3 = \frac{(250)^2}{R_3}
W1:W2:W3=15:25:64 or W1<W2<W3W_1 : W_2 : W_3 = 15 : 25 : 64 \ or \ W_1 < W_2 < W_3