SSC JE Electrical Engineering 2015 Paper

Question : 196 of 199

Marks: +1, -0

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

Solution:

V = IR

= 750 ×10^{–6} ×11 ×10^{–3}

= 8.25V

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