LED AND RESISTOR CIRCUIT (1)

Session 2, MLQ 2.4. Current can be measured across the resistor using V=IR, and Kirchoff's Second Law to calculate V: Battery voltage = 12V Voltage drop across an LED = 0.737V Voltage across R1 = (12 - 0.737). I = (12-0.737)/470 = 23.963mA. Note that, of course, when the voltage is positive, no current passes through LED2. When the voltage is reversed (i.e., is negative), no current passes through LED1. Which means that the modulus value of the voltage is the same, regardless of the direction of the voltage. Therefore, the modulus value of the current will be the same, regardless of the direction of the voltage. Whether the current is positive or negative will be the only difference in its value, and this will correspond to whether the voltage is positive or negative. So, for a positive voltage, the current will be positive; for a negative voltage, the current will be negative. In both cases, the modulus value of the current will be the same. [For interest: putting VA meters before and after the switch, showed the voltage drops to zero after the switch. The switch, therefore, must have some resistance (and is about 1 milliohm by calculation of V/I, where V = 2.3963 microvolts after the resistor R1 and before the switch.