# Physics 9702 Doubts | Help Page 191

__Question 928: [Current of Electricity]__
When four identical lamps P, Q, R
and S are connected as shown in diagram 1, they have normal brightness.

The four lamps and the battery are
then connected as shown in diagram 2.

Which statement is correct?

A The lamps do not light.

B The lamps are less bright than
normal.

C The lamps have normal brightness.

D The lamps are brighter than
normal.

**Reference:**

*Past Exam Paper – June 2002 Paper 1 Q34 & June 2006 Paper 1 Q36*

__Solution 928:__**Answer: C.**

Both diagrams consist of 2 loops in
parallel with the supply, each containing 2 lamps.

Diagram 2 can be re-drawn in a
similar way than that of diagram 1. In that case, lamps P and R would be in
series with each other in one branch, and lamps Q and S would be series with
each other in another branch.

Since all the lamps are identical,
the lamps would have normal brightness as the current flowing through them
would still be the same.

__Question 929: [Electric field]__
Electric field strength between a
pair of parallel plates is E. The separation of the plates is doubled and the
potential difference between the plates is increased by a factor of four.

What is the new electric field
strength?

A E B
2E C 4E D 8E

**Reference:**

*Past Exam Paper – June 2007 Paper 1 Q30*

__Solution 929:__**Answer: B.**

Electric field strength, E = V / d

where V is the p.d. between the
plates and d is the separation

The separation of the plates is
doubled (d becomes 2d) and the p.d is increased by a factor of 4 (V becomes 4V)

New Electric field strength = 4V /
2d = 2 (V / d) = 2E

__Question 930:__

__[Operational Amplifier]__**(a)**By reference to an amplifier, explain what is meant by

*negative feedback*.

**(b)**An amplifier circuit incorporating an ideal operational amplifier (op-amp) is shown in Fig.1.

Supply for the op-amp is ± 9.0 V.

The amplifier circuit is to have a
gain of 25.

Calculate the resistance of resistor
R.

**(c)**State the value of the output voltage V

_{OUT}of the amplifier in (b) for input voltages V

_{IN}of

(i) – 0.08 V,

(ii) +0.4 V.

**Reference:**

*Past Exam Paper – June 2009 Paper 4 Q10*

__Solution 930:__**(a)**Negative feedback occurs when (part of) the output is added to /returned to / mixed with the input and is out of phase with the input / fed to inverting input

**(b)**

{Equation for a
non-inverting amplifier: Voltage gain = V

_{OUT}/ V_{IN}= 1 + (R_{f}/ R_{1})}
25 =
1 + (120 / R)

Resistance
R = 5 kÎ©

**(c)**

(i)

{V

_{OUT}= Gain × V_{IN}= 25 × – 0.08 = – 2 V}
V

_{OUT}= –2 V
(ii)

{V

_{OUT}= Gain × V_{IN}= 25 × 0.4 = 10V. The voltage given out is greater than that supplied (9V). However, from the conservation of energy, this is not possible. So, the amplifier is saturated and the output voltage is thus 9V (the same as that supplied).}
V

_{OUT}= 9 V

__Question 931: [Current of Electricity]__
In the circuit shown, the 6.0 V
battery has negligible internal resistance. Resistors R

_{1}and R_{2}and voltmeter have resistance 100 kÎ©.
What is the current in the resistor
R

_{2}?
A 20 Î¼A B 30 Î¼A C
40 Î¼A D 60 Î¼A

**Reference:**

*Past Exam Paper – June 2008 Paper 1 Q37*

__Solution 931:__**Answer: A.**

Total resistance in circuit = 100 +
[1/100 + 1/100]

^{-1}= 100 + 50 = 150 kÎ©
Ohm’s law: V = IR

Total current I in circuit = 6.0 /
150 = 40 Î¼A

This current of 40 Î¼A flows through
resistor R

_{1}, but it splits at the junction of R_{2}and the voltmeter which are in parallel. Since both has the same resistance, the current splits equally.
Current in R

_{2}= 40 / 2 = 20 Î¼A
I had a question related to plotting points on the graph in especially P3 and P5...we usually get values that are,lets say, 3.4 small squares to the right of a value such as 3.4 small squares to the right of 2.5.

ReplyDeleteNow, the mark scheme says we should plot to within half a small square, does that mean that instead of plotting 3.4 small squares to the right of 2.5, we can plot 3.5 small squares to the right.

Similarly, if we have a value 4.8 small squares to the right of some other value, can we plot it at 5 small squares?(since it is quite difficult to plot to 3.4 and 4.8 small sq accurately).

Thank You.

Yes, but go to the 3.5 or the 5 small squares, and try to plot it a little close to it, either before or after. Because 3.5 would actually be another value. Just a little deviation. This would indicate the examiner that you actually wanted to plot at 3.4 or 4.8

DeleteWhat is most important is that you chose the proper scale for the axes. You should not obtain such cases if you choose a proper axis.

In the mark scheme its written that we should use scales which are multiples of 1,2 and 5 and not those which are multiples of 3,7,11 etc. Can you explain what is meant by this by an example because in practical work in P3, we usually obtain values such as 1.230 and 0.673 and so on .

DeleteIt means that the scale labels can be as follows:

Delete0, 1, 2, 3, 4, ....

OR 0, 2, 4, 6, 8, ...

OR 0, 5, 10, 15, 20, ...

but NOT

0, 3, 6, 9, 12, ...

OR 0, 7, 14, 21

The second set which is not allowed is difficult to read.

You just need to plot these points in the correct positions as accurately as possible - this may not always to exact. You may round to figures to an appropriate number of figures.

Please consider answering ALL of the following questions before October:

ReplyDelete4/O/N/02 Q.5(b),Q.6(c)(i)

6/O/N/02 Q.11(a)(b)

6/O/N/03 Q.9

04/M/J/04 Q.8(a),(b)(i),(ii)1.

06/M/J/04 Q.9(b)(iii),Q.11(b)

06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)

04/M/J/05 Q.7(a)

06/O/N/05 Q.8(b),Q.10(a)

04/M/J/06 Q.6(a),(c),Q.7(b)

06/M/J/06 Q.14(b)

04/O/N/06 Q.3(c)

06/O/N/06 Q.3(b)

05/M/J/07 Q.2(d)

04/O/N/07 Q.7(b)(i),(c),Q.10(c)

04/M/J/08 Q.3(c)(ii),Q.5(b),Q.9(b),Q.11(a)(ii)

04/O/N/08 Q.7(c)

04/M/J/09 Q.11(b)(iii)

41/O/N/09 Q.6(a),(b)(i),Q.10

42/O/N/09 Q.7(b)(ii)

41/M/J/10 Q.6(a),Q.7(a)

51/M/J/10 Q.2(d)

42/O/N/10 Q.3(c)

41/M/J/11 Q.8(a)

For 42/O/N/10 Q.3(c), go to

Deletehttp://physics-ref.blogspot.com/2014/09/9702-november-2010-paper-41-42-worked.html