A 2,V battery forward biases a diode. There i | vedclass

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(C) The circuit consists of a $2\,V$ battery,a diode,and a series resistor $R$ in a loop. Applying Kirchhoff's Voltage Law $(KVL)$ to the circuit: $V_

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$300\,\Omega$

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(C) The circuit consists of a $2\,V$ battery,a diode,and a series resistor $R$ in a loop. Applying Kirchhoff's Voltage Law $(KVL)$ to the circuit: $V_{battery} = V_{diode} + V_{resistor}$ Given $V_{battery} = 2\,V$,$V_{diode} = 0.5\,V$,and the desired current $I = 5\,mA = 5 	imes 10^{-3}\,A$. Substituting these values: $2\,V = 0.5\,V + I 	imes R$ $2 - 0.5 = 5 	imes 10^{-3} 	imes R$ $1.5 = 5 	imes 10^{-3} 	imes R$ $R = \frac{1.5}{5 	imes 10^{-3}} = \frac{1500}{5} = 300\,\Omega$ Therefore,the required series resistance is $300\,\Omega$.

$A$ $2\,V$ battery forward biases a diode. There is a drop of $0.5\,V$ across the diode which is independent of current. Also,a current greater than $10\,mA$ produces large joule heat and damages the diode. If the diode is to be operated at $5\,mA$,the series resistance to be used is:

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