1. (a) In the United States, the standard sinusoidal AC household voltage is 120

Question

1. (a) In the United States, the standard sinusoidal AC household voltage is 120 V (rms), and electric light bulbs are classified according to the rate at which electrical energy is consumed in them at that potential difference. In this problem we consider two old-style incandescent light bulbs, one of 60.0 W and the other of 100 W. (G) Calculate the resistance of each bulb, specifying clearly which is which. 14] (ii) What is the equivalent resistance of the bulbs connected in parallel? What peak current Imax will be supplied to the parallel combination by the 120-V rms source? (iii) If the bulbs are instead connected in series, what will Imax be now? In this case, the power consumed by the 100-W bulb will be how many times that of the 60-W bulb? (Give your answer as a fraction in lowest terms.) (b) A vacuum-filled parallel-plate capacitor has circular plates of diameter 12.0 cm and a capacitance of 4.38 F. It is connected in a circuit with a 2.00 H inductor, as shown in the figure at the right. The capacitor is charged to a potential difference of 12.0 volts, and then the switch S is closed at time t-0 C Taking the Maxwell-Ampère law into account, consider the magnitude of the induced magnetic field between the circular plates of the capacitor at the outer edge of the plates. Find the earliest time t>0 at which the magnetic field has its largest value, and calculate that value. = 8.85 x 10-12.) [12] (In SI units, ° = 4 x 10-7 and

Explanation / Answer

Dear student,

This is my first answer...

Answer for the first question

(i ) Incase of 60W bulb

Step 1 : power(p) = Voltage(V) * Current(I)

60= 120*I which gives (I) = 1/2 Amps

Step 2: voltage(v)= Resistance(R) * Current(I)

so,120 =R * (1/2)

which gives R = 240 Ohms This is the resistance of 60W bulb.

in the same way if u calculate for the 100W bulb ,you ll get 144 Ohms as answer.

(ii) Formulae for equalent resistance when 2 resiter connected in parallel is ,

= (R1 * R2) /(R1+R2) just apply R1 as 240 Ohms and R2 as 144 Ohms ( which is calculated earlier ),

The answer is 90 Ohms (Req) .

Peak current can be calculated by

I(rms) = Voltage (V) / Equalient resitance (Req)

I(rms) =120 / 90 = 1.333 Amps

Peak Current (I peak) = 1.414 * (I rms)

I(peak) = 1.414 & 1.333 = 1.8851 Amps

(iii) if bulbs are connected in series ,Equalient resistance will be addition of individual resistances

so R(eq) = R1 + R2 = 240+144= 384 Ohms.

I(rms) = 120 / 384= 0.312 Amps

again use same formulae for I(Peak) =1.414 * i(rms)

Peak current will be = 1.414 * .312= 0.4418 Amps

When th bulbs are connected in series ,current flowing through will be equal which is 0.312 Amps

According to the formulae P= (I)2 * R, current is equal so POWER CONSUMED BY THE BULBS IS PROPORTIONAL TO THEIR RESISTANCES.

The ratio of the resistance of the bulbs gives the ratio of power consumption, here

= 144 ohms( R of 100 W bulb ) / 240 Ohms ( R of 60 W bulb ) = 0.6 (Answer)

(B) Actually i am working on this i need some more time, to refer. At present i a unable to derive it.

I can answer if i get some more extra time . after referring my books i can answer this.

Thanks&Regards,

Ramesh

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