Two bulbs of 100, W, 220, V and 200, W, 220, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For the first bulb,$R_1 = \frac{220^2}{100} = 484\, \Omega$.For the second bulb,$R_2 = \f

Vedclass · Accepted Answer

$66$

Vedclass · Answer

(B) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For the first bulb,$R_1 = \frac{220^2}{100} = 484\, \Omega$.For the second bulb,$R_2 = \frac{220^2}{200} = 242\, \Omega$.When connected in series,the total resistance is $R_{eq} = R_1 + R_2 = 484 + 242 = 726\, \Omega$.The total power consumed is $P_{total} = \frac{V^2}{R_{eq}} = \frac{220^2}{726} = \frac{48400}{726} \approx 66.67\, W$.Alternatively,for series connection,$P_{consumed} = \frac{P_1 P_2}{P_1 + P_2} = \frac{100 	imes 200}{100 + 200} = \frac{20000}{300} = 66.67\, W \approx 66\, W$.

Two bulbs of $100\, W, 220\, V$ and $200\, W, 220\, V$ are connected in series to a $220\, V$ supply. What is the total power consumed in $W$?

Similar Questions

The equivalent resistance between the points $A$ and $D$ is ................ $\Omega$.

$A$ letter $'A'$ is constructed of a uniform wire with resistance $1.0\,\Omega/cm$. The sides of the letter are $20\,cm$ and the cross-piece in the middle is $10\,cm$ long. The apex angle is $60^{\circ}$. The resistance between the ends of the legs is close to ................ $\Omega$.

The figure shows a tetrahedron where each edge has a resistance $r$. Choose the correct statement$(s)$ related to the equivalent resistance between any two vertices.

The equivalent resistance across $A$ and $B$ as shown in the figure is ............. $\Omega$.

The resistance of the series combination of two identical resistors is $S$. When they are joined in parallel,the total resistance is $P$. If $S = nP$,then the minimum possible value of $n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module