In the figure below, what is the value ($$kOmega$$) of R needed to make $$i_1=1.25"mA"$$?

 

1 Answer

James Dowd

Updated on December 23rd, 2020

We first want to simplify the circuit to find the equivalent resistance (Req).

Resistors in series add together, so we can add the two 4kΩ resistors together.

Req1=4kΩ+4kΩ=8kΩ

Next, we want to get an expression for the combined parallel resistance of Req1 and R. We can use the equation for finding the equivalent resistance of two resistors in parallel to do so:

Req=R1R2R1+R2 Req2=(R)(8kΩ)R+8kΩ

Finally, we add this to the 4kΩ resistor in series to find the equivalent resistance of the circuit.

Req=4kΩ+Req2 Req=4kΩ+(R)(8kΩ)R+8kΩ Req=(4kΩ)(R)+(32000kΩ)R+8kΩ+(R)(8kΩ)R+8kΩ Req=(12kΩ)(R)+(32000kΩ)R+8kΩ 

Now that we have the equivalent resistance, we use Ohm's law to solve for R:

V = IR 10V= 1.25mA *((12kΩ)(R)+(32000kΩ)R+8kΩ) 8000 = (12k)(R)+32000kR+8k 8kR + 64000k = 12kR + 32000k 4kR = 32000k R = 8kΩ

Therefore, R must be equal to 8kΩ.

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