• Physics 9702 Doubts | Help Page 45

Question 271: [Circular motion]

(a) Define the radian

(b) Stone of weight 3.0 N is fixed, using glue, to one end P of rigid rod CP, as shown.

The rod is rotated about end C so that the stone moves in vertical circle of radius 85 cm.

Angular speed ω of rod and stone is gradually increased from zero until the glue snaps. The glue fixing the stone snaps when tension in it is 18 N.

For position of the stone at which the glue snaps,

(i) on dotted circle of Fig, mark with the letter S the position of the stone

(ii) calculate angular speed ω of the stone.

Reference: Past Exam Paper – June 2010 Paper 41 Q1



Solution 271:

Go to
A stone of weight 3.0 N is fixed, using glue, to one end P of a rigid rod CP, as shown  in Fig. 1.1.

Question 272: [Current of Electricity > Capacitance]

Rectified output of sinusoidal signal generator is connected across a resistor R of resistance 1.5 kΩ, as shown.

Variation with time t of potential difference V across R is shown.

(a) State how rectification shown in Fig may be achieved.

(b) A capacitor is now connected in parallel with resistor R. Resulting variation with time t of potential difference V across R is shown.

(i) Using Fig, determine

1. mean potential difference across resistor R

2. mean current in resistor

3. time in each cycle during which capacitor discharges through the resistor.

(ii) Using answers in (i), calculate

1. charge passing through the resistor during one discharge of the capacitor

2. capacitance of capacitor

(c) Second capacitor is now connected in parallel with resistor R and the first capacitor. On Fig, draw a line to show variation with time t of the potential difference V across the resistor.

Reference: Past Exam Paper – November 2003 Paper 4 Q4



Solution 272:

(a) By connecting a single diode EITHER in series with resistor R OR in series with an a.c. supply

(b)

(i)

1.

{Maximum V = 6.0V. Minimum V = 4.8V. Mean potential difference = (6.0 + 4.8) / 2 = 5.4V}

(From graph,) 5.4 V

2.

Ohm’s law: V = IR

Current I = 5.4 / (1.5x10³) = 3.6x10^-3A

3.

{As V falls, the capacitor discharges by supplying charge. So, the cycle during which the capacitor discharges through the resistor corresponds to the each region where V decreases.}

Time = 0.027s

(ii)

1.

Charge Q = IT = (3.6x10^-3) x 0.027 = 9.72x10^-5C

2.

{Since the CHANGE in the charge has been calculated, we need to substitute the change in potential difference into the formula for capacitance. Change in V = 6.0 – 4.0 = 1.2V}

Capacitance C = ΔQ / ΔV = (9.72x10^-5) / 1.2 = 8.1x10^-5F

(c)

{If no capacitor was present (capacitance = 0), then a graph similar to part (a) is obtained.  So, by increasing the capacitance, the decrease in V is less – we have the smoothing effect. So, a greater value of capacitance causes a greater smoothing effect. Connecting capacitors in parallel causes the total capacitance in the circuit to increase. Thus, in the graph, the ripple should be less.}

The line should have a reasonable shape with less ripple

Question 273: [Current of Electricity > Resistance]

What is the equivalent resistance of this circuit?

Solution 273:

First label the different junctions present in the circuit. Note the junction A and B are the same and E and F are the same.

The 1Ω resistor is between B and C but considering the upper part, junction B is the same as D (no component is connected in the upper wire connecting B and D). This means that the 1Ω resistor (currently between B and C) can also be drawn between C and D. So the circuit can also be drawn as:

Similarly, considering the bottom part between C and E (they are the same as no component is connected between them), the circuit becomes

So, the equivalent resistance = [1/1 + ½ + 1/3]^-1 = 6/11

Question 274: [Current of Electricity > Resistance]

Circuit contains three similar lamps A, B and C. Circuit also contains three switches, S₁, S₂ and S₃, as shown.

One of the lamps is faulty. In order to detect fault, an ohm-meter (meter that measures resistance) is connected between terminals X and Y. When measuring resistance, ohmmeter causes negligible current in the circuit.

Fig shows readings of the ohm-meter for different switch positions.

(a) Identify faulty lamp, and nature of the fault

(b) Suggest why it is advisable to test the circuit using an ohm-meter that causes negligible current rather than with power supply.

(c) Determine resistance of one of the non-faulty lamps, as measured using the ohmmeter.

(d) Each lamp is marked 6.0 V, 0.20 A.

Calculate, for one of the lamps operating at normal brightness,

(i) its resistance

(ii) its power dissipation

(e) Comment on answers to (c) and (d)(i).

 

Reference: Past Exam Paper – June 2006 Paper 2 Q7

Solution 274:

(a)

{When all the switches are open, the meter reads infinity. This is obvious since the circuit is shorted.

In the third test: S₁: open, S₂: closed and S₃: open, the meter read a resistance of 30Ω. This is because test 2, where only S₁ is closed, proves that lamp A is NOT faulty. So, the third test proves that lamp B is also NOT faulty.

In the last test, where only S₁ is opened, the resistance read is 15Ω. This proves that lamp C is shorted – that is current passes as if there was no lamp. In this case, it’s as if a wire of negligible resistance is connected in parallel with lamp B.

Current is larger when the resistance is smaller. [From Ohm’s law, current I is inversely proportional to resistance R] So, all current would flow in that wire instead of the lamp B (since lamp B has some resistance). So, the only resistance encountered by the current is due to lamp A only. So, the resistance read by the meter is 15Ω. 

If S₁ and S₂ were closed, and S₃ was opened, the meter would again read 15Ω since all the current would pass through S₁ (which has no resistance) instead of passing through lamp B. The current would then flow through lamp A (which is the only lamp that causes a resistance in this case).} 

Faulty lamp: Lamp C

Nature of fault: The lamp is shorted

(b) The shorted lamp A would cause damage to the supply/lamps / blow fuse in the supply

(c) Resistance = 15Ω

(d)

(i)

V = IR

Resistance R (= V / R = 6.0 / 0.20) = 30Ω

(ii)

Power P = VI  or I²R  or V²/R

Power P = 1.2W

(e)

The filament is cold when measuring with the ohm-meter in (b). The resistance of the filament rises as the temperature rises.