# Help understanding combination series/parallel circuits

Discussion in 'General Electronics Chat' started by zogdc, Sep 16, 2013.

1. ### zogdc Thread Starter New Member

Sep 16, 2013
1
0
I'm very new to electronics and am playing around with different combinations of resistors and LEDs trying to master the concepts behind series and parallel circuits. For instance, creating circuits with known resistance values on a bread board, then calculating voltage and current and different points in the circuit, then using a multimeter to see if the real world observed values match my calculations.

The circuit below has me stumped, and I'm not sure how to approach it. I begin by calculating the combined resistance of the parallel section of the circuit as:

1 / (1/220 + 1/470) ~= 150 Ω

and total resistance on the circuit is therefore 620 Ω.

Voltage drop across R1 should be 470/620 * 9 = 6.8v, and voltage drop across the parallel section should be 2.2v. This is where I think I go wrong because my calculations for each branch of the parallel circuit aren't internally consistent. Also my observed values are more like 5.7v and 3.3v.

Can anyone help?

File size:
19.4 KB
Views:
45
2. ### aamir_uetn New Member

Jul 26, 2013
2
0
only one problem, u didnt include LED's resistance , it should b around 110 ohms if drop across LED is 2.2V

3. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,991
1,115
Simply replace the LED diode with a 2.2V voltage source. And repeat your calculations.
Do you know Thevenin's theorem ?

4. ### #12 Expert

Nov 30, 2010
16,685
7,324
I did that circuit yesterday in the Homework section, except the parallel load was a transistor with a gain of 80. This is probably homework, too.

You have to use the voltage across the LED, not its alleged or apparent resistance. Set up an equation where (the sum of the branch currents) = (9V -the voltage after the first resistor)/R1

After that, it's all algebra.