Could someone please help me to calculate the output voltage across the voltmeter of a resistor divider network as shown in below schematic whose input can vary from -20V to +20V, R1 = 43K, R2 = 5.1K and Bias Voltage on R2 is 2.5V.


simulate this circuit – Schematic created using CircuitLab


Handy formula, and it works a treat!

enter image description here


Forget about VM1, If we simplified the circuit and connect two Voltage source of 20 V and 2.5 V, by connecting their GND, over all voltage should be 20V - 2.5V = 17.5 V, now two resistor in series, so over all current trough circuit is, I = V / R. where R = 43 + 5.1 = 48.1 kOhm, and V = 17.5 V so I = 17.5 V / 48.1 kOhm = 0.363825364 mA, Now as you can see, voltage drop across R2 is,

Voltage drop across V_R2 = R2 x I = 1.855509356 V and finally Volt Meter will read value Vm1 = V_R2 + V1 = 2.5 + 1.855509356 = 4.355509356 V

Is that clear?

  • 1
    \$\begingroup\$ He's looking for the general formula to cover V2 from -20 V to +20 V. You have only provided a specific answer at one input voltage. "k" for kilo. "K" is for kelvin. 9 decimal places on your calculations is probably a bit silly given practical resistor tolerances. \$\endgroup\$ – Transistor Mar 27 '16 at 15:25
  • \$\begingroup\$ oh yes... i updated for that \$\endgroup\$ – Prakash Darji Mar 27 '16 at 15:54

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