1992 paper1 (more pain and misery )
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Re: 1992 paper1 (more pain and misery )
paper 2 q5 tryed to use y=mx+c but cannot take exact points
ANDYQ- Posts: 23
Join date: 2009-02-19
Re: 1992 paper1 (more pain and misery )
won,t be on to around 7pm tommorrow got a 9h shift at work (goin for the sympathy vote guys)
ANDYQ- Posts: 23
Join date: 2009-02-19
Re: 1992 paper1 (more pain and misery )
Paper I Q34 Andy the loop of wire has resistance 30 ohms, it is connected to the4 circuit at the midpoint, that gives you 2 paths for the current to flow each has a resistane of 15 ohms. To solve the question find the resistance of a circuit that has two 15 ohm resistors in parallel.
Paper II Q 5 Take the following points (0,2.0) and (0.5,1.2) this makes a gradient of -1.6 so internal resistance = 1.6 ohms.
Paper II Q 5 Take the following points (0,2.0) and (0.5,1.2) this makes a gradient of -1.6 so internal resistance = 1.6 ohms.

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Re: 1992 paper1 (more pain and misery )
5.a.i. Draw the circuit given, add an ammeter put the variable resistor between the output terminals and put a voltmeter across the variable resistor. Set the variable resistor to a resistance, press a key and record both meter readings in a table. Change the resistance of the variable resistor and repeat the experiment. Collect 5 sets of data and plot in a graph.
5.b. The reading from figure 1 gives you the EMF (E), the reading on figure 2 gives you the voltage across the variable resistor. Because we know R and V in figure 2 we can work out I, I = V / R. Then using the equation E = IR + Ir where we know E, I and R we get r = (E- IR) / I.
8.b.ii. Clue is question "the pads are picking up a sinal from mains wiring" in the inverting mode this signal will be displayed on the monitor, in the differential mode the signal from 1 pad at the positive input will be cancelled by the same signal going into the negative input.
5.b. The reading from figure 1 gives you the EMF (E), the reading on figure 2 gives you the voltage across the variable resistor. Because we know R and V in figure 2 we can work out I, I = V / R. Then using the equation E = IR + Ir where we know E, I and R we get r = (E- IR) / I.
8.b.ii. Clue is question "the pads are picking up a sinal from mains wiring" in the inverting mode this signal will be displayed on the monitor, in the differential mode the signal from 1 pad at the positive input will be cancelled by the same signal going into the negative input.

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