Wednesday, March 4, 2015

Physics 9702 Doubts | Help Page 76

  • Physics 9702 Doubts | Help Page 76



Question 406: [Current of Electricity]
The resistance of a thermistor decreases significantly as its temperature increases.
Which graph best represents the way in which the current I in the thermistor depends upon the potential difference V across it?
 
Reference: Past Exam Paper – J97 / I / 14 & N2000 / I / 13



Solution 406:
Answer: C.
Ohm’s law: V = IR
Current I = V / R
{An I-V graph for the above equation would be a straight line with positive gradient.}

But since resistance R decreases as the potential difference V increases, the values of current I would be greater than those on the straight line as V increases.









Question 407: [Current of Electricity > Internal resistance]
A cell is connected to resistor.

At any given moment, potential difference across the cell is less than its electromotive force.
Which statement explains this?
A The cell is continually discharging.
B The connecting wire has some resistance.
C Energy is needed to drive charge through the cell.
D Power is used when there is a current in the resistor.

Reference: Past Exam Paper – November 2009 Paper 11 Q30 & Paper 12 Q29



Solution 407:
Answer: C.
Energy is needed to drive charge through cell. That is, the cell has some internal resistance.

The questions states that the potential difference across the CELL is less than its electromotive force. So, the wires and resistor has nothing to do with it. [B and D are incorrect]










Question 408: [Current of Electricity]
Source of e.m.f. of 9.0 mV has internal resistance of 6.0 Ω.
It is connected across galvanometer of resistance 30 Ω.
What will be the current in galvanometer?
A 250 μA                    B 300 μA                    C 1.5 mA                    D 2.5 mA

Reference: Past Exam Paper – June 2010 Paper 11 Q35 & Paper 12 Q33 & Paper 13 Q31



Solution 408:
Answer: A.
Total resistance R in circuit = 6.0 + 30 = 36 Ω

Ohm’s law: V = IR
Current I = V / R = (9.0x10-3) / (6 + 30) = 2.5x10-4A = 250x10-6A = 250μA









Question 409: [Current of Electricity > Potential divider]
Potential divider circuit consists of 2 resistors of resistances P and Q, as shown. 


Battery has e.m.f. E and negligible internal resistance.
(a) Deduce that potential difference V across resistor of resistance P given by V = [P / (P+Q)] E.

(b) Resistances P and Q are 2000 Ω and 5000 Ω respectively. Voltmeter connected in parallel with 2000 Ω resistor and thermistor connected in parallel with 5000 Ω resistor, as shown. 

Battery has e.m.f. 6.0V. Voltmeter has infinite resistance.
(i) Explain qualitatively the change in reading of voltmeter as temperature of thermistor is raised.
(ii) Voltmeter reads 3.6 V when temperature of thermistor is 19°C. Resistance of thermistor at 19°C.

Reference: Past Exam Paper – November 2008 Paper 2 Q7



Solution 409:
(a)
EITHER
V = IP
Current in the circuit = E / (P+Q)
Hence, V = EP / (P+Q)

OR
The current is the same throughout the circuit.
V / P = E / (P+Q)
Hence, V = EP / (P+Q))

(b)
(i)
(As the temperature rises,) the resistance (of the thermistor) decreases.
EITHER The resistance of the parallel combination decreases OR the potential difference across 5kΩ resistor / thermistor decreases.
So, the potential difference across 2000Ω resistor / voltmeter reading increases.

(ii)
Let R be the resistance of the parallel combination,
{From the potential divider equation in the form: V = EP / (P+Q),}
EITHER 3.6 = (2 x 6) / (2+R)
Resistance R = 1.33kΩ
(T: resistance of thermistor)
1 / 1.33 = (1/5) + (1/T)
T = 1.82kΩ

OR
Current in 2kΩ resistor = (3.6V / 2000Ω =) 1.8mA
Current in 5kΩ resistor = ({6.0 V – 3.6V} / 5000Ω =) 0.48mA
Current in thermistor = (1.8 – 0.48 =) 1.32mA
{Potential difference across parallel combination of 5kΩ resistor and thermistor: 6 – 3.6 = 2.4V}
T = 2.4 / 1.32 = 1.82kΩ




12 comments:

  1. in q408, why is the resistance taken to be 36, shouldn't it be 5 ohms(1/R=1/30+1/6=1/5)?

    ReplyDelete
    Replies
    1. The internal resistance of the battery is 'inside' the battery. So, the connection is a series connection. With only a battery and a galvanometer, there cannot be a parallel connection.

      Delete
    2. But doesn't it say "it is connected ACROSS a galvanometer" so isn't it actually a parallel connection.

      Delete
    3. There;s only 2 components. There is no way that they can be connected in parallel to form a complete circuit.

      Try to draw them in parallel. You would see that they will be in series with each other.

      Delete
    4. Well that was silly of me, thank you.

      Delete
  2. In solution 409 why are we subtracting the currents for getting the current in the thermistor?

    ReplyDelete
    Replies
    1. The 5kΩ resistor and the thermistor are in parallel with each other. Current splits at a junction of 2 components.

      The total current in the circuit is the same current flowing through the 2kΩ resistor. By finding the current flowing in the 5kΩ resistor, we can determine the remaining current that went to the thermistor.

      Delete
  3. hello in soloution 409 ii why is 1/1.33 = to 1/5 + 1/T and why is it not just 1.33 is equals to it

    ReplyDelete
    Replies
    1. it's the formula for parallel combination.

      1/R = 1/R1 + 1/2R2

      Delete

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