If two bulbs of power 25 W and 100 W respectively each rated at 220 V are connected in series with the supply

Q: If two bulbs of power 25 W and 100 W respectively each rated at 220 V are connected in series with the supply of 440 V. Which bulb will fuse?

(a) 25 W bulb

(b) 100 W bulb

(c) none of these

(d) both

Ans: (a)

Sol:
Resistance of one bulb, R1 = (220)2/25 Ω ;

Resistance of other bulb, R2 = (220)2/100 Ω.

Total resistance when two bulbs are in series,

R = (220)2/25 + (220)2/100 = (220)2/20

current I = V/R = 440/((220)2/20) = 2/11 amp.

Pot. diff. across 25 watt bulb
= IR1 = 2/11 × (220)2/25 = 352 V

Pot. diff. across 100 watt bulb
= IR2 = 2/11 × (220)2/100 = 88 V

Thus, the bulb 25 W will be fused, because it can tolerate only 220 V while the voltage across it is 352 V.

A and B are two points on a uniform ring of resistance R. The ∠AOB = θ…

Q: A and B are two points on a uniform ring of resistance R. The ∠AOB = θ, where O is the center of the ring. The equivalent resistance between A and B is:

(a) \displaystyle \frac{R \theta}{2 \pi}

(b) \displaystyle \frac{R (2\pi - \theta)}{4 \pi}

(c) \displaystyle R (1- \frac{\theta}{2 \pi} )

(d) \displaystyle \frac{R}{4 \pi^2} (2\pi -\theta )\theta

Ans: (d)

Sol: Let r be the radius of the ring.

Resistance per unit length of ring , ρ = R/(2 π r)

Lengths of section making angle θ = r θ

Resistance of this section \displaystyle R_1 = \rho r \theta = \frac{R}{2 \pi r} r \theta

\displaystyle R_1 =  \frac{R \theta}{2 \pi }

Lengths of section making angle (2π -θ) = r (2π -θ)

Resistance of this section \displaystyle R_2 = \rho r (2\pi -\theta) = \frac{R}{2 \pi r} r (2\pi -\theta)

\displaystyle R_2 =  \frac{R (2\pi - \theta ) }{2 \pi }

Here, R1 and R2 are in parallel between A and B,

\displaystyle R_{eq} = \frac{R_1 R_2 }{R_1 + R_2}

\displaystyle R_{eq} = \frac{\frac{R \theta}{2 \pi }\times  \frac{R (2\pi - \theta )}{2 \pi} }{\frac{R \theta}{2 \pi } + \frac{R (2\pi - \theta ) }{2 \pi }}

\displaystyle = \frac{R}{4 \pi^2} (2\pi -\theta )\theta

Masses of the three wires of same material are in the ratio of 1 : 2 : 3 and their lengths in the ratio…

Q: Masses of the three wires of same material are in the ratio of 1 : 2 : 3 and their lengths in the ratio of 3: 2: 1, Electrical resistance of these wires will be in the ratio of:

(a) 1 : 1: 1

(b) 1: 2: 3

(c) 9: 4: 1

(d) 27: 6: 1

Ans: (d)

Sol : Mass, M = volume × density = A l × d

A = M/ld

Resistance \displaystyle R = \rho \frac{l}{A} = \frac{\rho l}{M/ld}

\displaystyle R = \rho \frac{l^2 d}{M}

\displaystyle R = \propto \frac{l^2 }{M}

\displaystyle R_1 : R_2 : R_3 =  \frac{l_1^2 }{M_1} : \frac{l_2^2 }{M_2} : \frac{l_3^2 }{M_3}

\displaystyle  =  \frac{3^2 }{1} : \frac{2^2 }{2} : \frac{l^2 }{3}

= 27 : 6 : 1