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1. 1)An LED connected in series with a resistor to a 5V supply. Calculate the resistance of the series resistor required to give a current in the LED of 12 mA and a voltage across it of 2 V?
2)A potential divider has a total resistance of 1 kiloohms and carries a current of 3mA. What is the output p.d. if the sliding contact id at its centre point?
3a)A 9 volt battery has an internal resistance of 12 ohms.
What is the potential difference across its terminals when it is supplying a current of 50mA.
b)what is the maximum current which the battery could supply?
2. Well for number 1.

In series, Voltage is shared and current is the same. So 12*10^-3 Amps is the same for the resistor as well as the LED.Since voltage is shared.. 3V goes through the Resistor (since 2V goes through the LED).
So now we can find the resistance of the resistor... R = V/I therefore R=3/12*10^-3. = 250 ohms.

number 2. we can work out the overall voltage... (1*10^3) * (3 *10^-3) = 3 Volts. When the slider is at the centre point the resistance is half for each "resistor" in the potential divider...so resistance is the same for each resistor so the ratio is a half... so the answer is 1.5V

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