# Voltage Drop on parallel resistor

I am trying to verify voltage drop across a resistor as seen on my multimeter with basic ohms law calculations.

As shown in below circuit (from Electronics for DUMMIES). The multimeter shows a voltage drop of 6V at 1M ohm resistor. However that is not what my math shows and I think something is wrong with my maths. Here is how I am doing it

Vb=9V
First Find total Resistance of circuit
R1=1 M ohms= 1000000 ohms
R2=10 k ohms = 10000ohms
R3=470 ohms

R1 and R2 are in series with each other but both combined are in parallel to R3, so
Rtotal = (R1+R2)||R3 = (1000000+10000)||470
= 1010000 || 470
= (1010000 x 470)/(1010000 + 470)
=474700000/1010470
=469.7813888586499=469.7

Now to find Total Current
I=9V/469.7ohms
= 0.019 Amps

Voltage drop across  R1(1 M ohms)
V1=0.019 x 1000000 = 19000(This cant be right)


Did I miss anything?

The NPN transistor is BC546B, which is acting as an amplifier

• Did you take into account the meter has a resistance value which effects the voltage drop across the 1M0 by lowering the overall resistance (gives a lower reading)? – JIm Dearden Apr 22 '16 at 19:03
• Ditto and how about telling us what R1, 2 and 3 are in your circuit. Also how are R1 + R2 in parallel with R3? – Andy aka Apr 22 '16 at 19:05
• yea I did, but still the calculation seems way off – user482963 Apr 23 '16 at 7:58