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1830818186's Question
Physics E&M
Posted about 2 months ago

The circuit shown in the figure employs an operational amplifier that can be considered ideal. Determine the value of the current I0I_{0} delivered to the load resistance RLR_{L}.
Data: R1=R2=2[kΩ],RL=1[kΩ],VN=5[V]R_{1}=R_{2}=2[k \Omega], R_{L}=1[k \Omega], V_{\mathbb{N}}=-5[V].
(Glossary: V=\mathrm{V}= volt, m=\mathrm{m}= meter, A=\mathrm{A}= ampere, W=\mathrm{W}= watt, Ω=\Omega= ohm, F=\mathrm{F}= farad, Hz=\mathrm{Hz}= hertz, s=\mathrm{s}= second)

Select one:
a. I0=5[ mA]I_{0}=-5[\mathrm{~mA}]
b. I0=5[ mA]I_{0}=5[\mathrm{~mA}]
c. I0=1[ mA]I_{0}=1[\mathrm{~mA}]
d. I0=1[ mA]\mathrm{I}_{0}=-1[\mathrm{~mA}]
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Answer from Sia
Posted about 2 months ago
D
Key Concept
Inverting Operational Amplifier
Explanation
In an inverting operational amplifier, the output voltage is given by Vo=(R2R1)VinV_o = -\left(\frac{R_2}{R_1}\right) V_{in}. Given R1=R2=2kΩR_1 = R_2 = 2 \, k\Omega and Vin=5VV_{in} = -5 \, V, the output voltage Vo=(22)(5)=5VV_o = -\left(\frac{2}{2}\right) (-5) = 5 \, V. The current through the load resistor RL=1kΩR_L = 1 \, k\Omega is I0=VoRL=5V1kΩ=5mAI_0 = \frac{V_o}{R_L} = \frac{5 \, V}{1 \, k\Omega} = 5 \, mA. Since the current direction is opposite to the assumed direction, I0=5mAI_0 = -5 \, mA.

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