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14_faq

Page history last edited by johnlittlephysics 9 years, 10 months ago

14 D.C. Circuits FAQ


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Comments (7)

Goh Ying Ting said

at 1:38 pm on Jun 5, 2014

Hi Mr Ang, I would like to ask how do we determine the trend of resistance (i.e. resistance is increasing at a decreasing rate etc) for iv graphs or vi graphs? For instance, for the iv graph of a thermistor on page 8 of the notes (qn 1), does the resistance increase at a decreasing rate? Thank you!

johnlittlephysics said

at 5:56 pm on Jun 5, 2014

From the IV or VI graph of a device, we cannot tell if the resistance is changing at an increasing or decreasing rate with increase in V or I. Just like the question in the recent Sec 4 MYE Paper 2 graph, we can only tell if the resistance is increasing or decreasing, not the rate of change.

For the shape of the IV graph of the thermistor (14.3 Q1), we can only say the resistance R decreases with increase in V, not the rate of decrease.
Note: saying how R changes, is like describing how the slope of the graph changes (although R is strictly a ratio, not gradient).

johnlittlephysics said

at 5:23 pm on Jun 5, 2014

Kah Leng: I would like to ask you a few questions regarding the exercise in the DC circuit notes:
1. Could you explain the answer for potential divider circuit 4(b)?
2. For light-dependent resistor 2(b) and (c), the answer state that the pd across the bulbs are "high enough" for the bulb to light up. How does one know or determine whether the pd is "high enough"? Because the question does not state the rating of the bulb...

johnlittlephysics said

at 6:05 pm on Jun 5, 2014

1. 14.2 Q4(b):
Similar to Q4(a), determine the p.d. Vo across variable resistor using potential divider method.
Vo = [2.0/(10 + 2.0)] x 24.0 V = 4.0 V.

Vo = potential difference between potential at P (Vp) and the potential at 24.0V (higher potential)
4.0 = 24.0 - Vp => Vp = 24.0 - 4.0 = 20.0 V

johnlittlephysics said

at 6:12 pm on Jun 5, 2014

2. 14.4 LDR Q2(b):
The question already states that the light bulb lights up automatically in the dark, and calculations in (a) shows that this occurs when the p.d. across the light bulb is 8.1 V (when LDR is in the dark). Hence we can conclude that 8.1 V is high enough to turn on the light bulb.
Simillarly for Q2(c), when the p.d. across the light bulb is 11.9 V (greater than 8.1 V), the light bulb would light up.

Goh Ying Ting said

at 11:33 am on Jun 14, 2014

Hi Mr Ang, I would like to ask what does "two lamps are first connected in series and then in parallel with a battery" mean? (textbook qn 9b pg 373) Thank you! :)

johnlittlephysics said

at 10:33 pm on Jun 16, 2014

There are two cases:
1. Two lamps are connected in series with the battery: so each has p.d. 6.0/2 = 3.0 V (assuming the lamps are identical)
2. Two lamps are connected in parallel with each other and with the battery: so p.d. across each lamp = e.m.f. of battery

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