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I replied in the other threadHey guys can we discuss chemistry paper 5 please.. as in question paper? Can anyone please help me with 2014 winter paper 52 question 2 (e,f) please? Thank you
but the resistance of thermistor was 120 and increased by 121.5.so wont it be 121. + 120/341.5You use the potential divider formula. For 121.5 ohms, the total voltage in the first segment is (121.5/121.5 + 120)*2000mV which gives you like 1006mV and then for the second part of the circuit (5000/5000+5000)*2000mV which gives you 1000mV. So 1006-1000 = 6mV
Why would it be 341.5?but the resistance of thermistor was 120 and increased by 121.5.so wont it be 121. + 120/341.5
same value for pd but i marked lowest point with M, dont know whyM at the top right, not sure if that is correct. And pd was 6mV.
I did too but then I changed it.same value for pd but i marked lowest point with M, dont know why
Inverse to this, 0.618. Output can't be more than input as there was attenuation.what was the answer to q11 part c, ratio P out/P in? i got the answer 1.62:/
2.09 was attenuation... 2.09=10log(out/in), wasn't this the method to find it?I did too but then I changed it.
Inverse to this, 0.618. Output can't be more than input as there was attenuation.
You wrote the formula for Gain which is =10log(Pout/Pin)2.09 was attenuation... 2.09=10log(out/in), wasn't this the method to find it?
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