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Meter Galvanometer
I need help in this question about electricity!?
AB is one metre length of uniform resistance wire connected in series with a 2v accumulator of negligible internal resistance. A centre reading galvanometer G in series with cell X ot Y , are connected across part of the resistance wire, as shown here : http://yfrog.com/0hcircuitbj
The setup is used to compare the EMFs EX and EY of two cells X and Y of nominal value 1.5V.
i) With no cell X or Y connected in the circuit, find the pd drop per cm of the resistance wire.
ii)The two cells X and Y give balance lengths of 764mm and 746mm respectively , measured from side B , with G registering no current. Find the ratio EX/EY.
iii) Explain why the internal resistance of the cells X and Y has no effect on this ratio.
Can someone help me please ? Will give Best Answer !
http://yfrog.com/0hcircuitbj
AB = 1meter.
P.D across AB = 1.5V
1)
With no cell X or Y connected in the circuit, the pd per cm of the resistance wire. is 1.5 volt /100 cm = 0.015 V/cm.
-------------------------
2)
EX = 0.015 V/cm *764
EY = 0.015 V/cm *746
EX/EY= 764/746 = 1.024
iii)
Since the potential differences are balanced , there is no current flowing through the cells Xor Y.
Since there is zero current , there is no potential.drop across the internal resistance and has no effect on this ratio.
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half deflection
Electrical Meters?
A particular galvanometer requires a current of 1.50 mA for full-scale deflection and has a resistance of 75.0. It can be used to measure voltage by wiring a large resistor in series with the galvanometer. The effect is to limit the current in the galvanometer when large voltages are applied. Calculate the value of the resistor that allows the galvanometer to measure an applied voltage of 25V at full-scale deflection.
Thank u!
16.7 K-ohms
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