# Help with question 6 of worksheets (Kirchhoff's Laws)

Discussion in 'Homework Help' started by yoflippo, Apr 19, 2010.

1. ### yoflippo Thread Starter New Member

Apr 19, 2010
3
0
Hi,

I can't figure out how question 6 has to be calculated.
This is the link to the question

I started with calculating the total resistance and got 0.009 [A].
Then I tried to calculate the current through the 1k+1k5+690 resistors.
The answer I found was 0,0054 [A].
So I also know the current through resistor 4700 [O] -> 0,0036 [A].
By calculating the voltage over the resistor 250 [O] I got: 0,009*250=2,25 [V]
By calculating the voltage over resistor 690 [O] I got: 0,0054*690=3,726 [V].
Summing the last to Voltages, I get: 5,976 [V]

But the correct answer should be 6,148 Volt.

Does anyone know what I'm doing wrong?

Last edited: Apr 19, 2010
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
The R_total is equal 2.15K
So I_total = 9.3mA
The voltage on 250 resistor is equal V_A = -9.3mA*250 = -2.325V
The current through the 1k+1k5+690 resistors is equal
I = I_total - (20V - 2.325V)/4.7K = 9.3mA - 3.76mA = 5.54mA
So voltage drop on a 690 resistor is equal
V_B = 5.54mA*690 = 3.82V
So V_AB = -2.325V - 3.822V = -6.147V

3. ### yoflippo Thread Starter New Member

Apr 19, 2010
3
0
Hi Jony130,

Thank you for your reply. Thanks to you I now see I have to adjust my calculating accuracy.

Best regards,

Yoflippo