Observations: when the cable was placed in both steps 2 and 3 the LED got bright

Question

Observations: when the cable was placed in both steps 2 and 3 the LED got brighter

1. Is the difference in net resistance in each case consistent with the brightness of the LED?

2. Calculate the equivalent resistance in the series and parallel cases. Based on this calculation, estimate the amount of current flowing through the diode in these two cases. Please show all of your work.

Procedure 1, Create the snap circuit depicted in Figure 5 using both the 1 and the 100 resistors, along with an LED. Note the brightness of the LED using this configuration. Figure 5: Initial configuration for Experiment 1 Connect the jumper wire to the point A marked in the photo. Note what happens to the LED when you connect and disconnect this wire. Reconfigure the circuit to that pictured in Figure 6. Turn on the switch and note the brightness of the LED 2. 3. Figure 6: New configuration for Experiment 1

Explanation / Answer

1) Yes. The difference in net resistance in step 2 and 3 is consistent with the brightness of the LED.

2) Series equivalent resistance RSE = 1000+100 = 1100 ohms.

The current flowing through the diode is ISE = V/1100 Amps where V is the battery voltage.

Parallel equivalent resistance RP = 1000*100/1100 = 1000/11 ohms

The current flowing through the diode is IP = 11V/1000 Amps

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