Question

If 750 µa is flowing through 11 k of resistance, what is the voltage drop across the resistor?


a. 8.25 v


b. 82.5 v


c. 14.6 v


d. 146 v

Answer 1

Current, I = 750 µA = 7.5×10^(-4) A

Resistance, R = 11000 ohm

Voltage drop, V = IR = 7.5×10^(-4) × 11000 = 8.25 Volt

Voltage drop is 8.25 volt.

Answer 2

Ohm’s principal discovery was that the amount of electric current through a metal conductor in a circuit is directly proportional to the voltage impressed across it, for any given temperature. Ohm expressed his discovery in the form of a simple equation, describing how voltage, current, and resistance interrelate:

$V=IR$

In this algebraic expression, voltage (V) is equal to current (I) multiplied by resistance (R). Using algebra techniques, we can manipulate this equation into two variations, solving for I and R, respectively:

$I=\frac{V}{R}$  $R=\frac{V}{I}$

Current, $I = 750 \mu A = 7.5\times 10^{-4} A$

Resistance, $R = 11000 ohm$

Voltage drop, $V = IR = 7.5\times 10^{-4}\times 11000 = 8.25 Volt$

Voltage drop is $8.25$ volt.

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