A diode having a potential difference of 0.5, | vedclass

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(C) The diode and the resistor are connected in series with the voltage source.According to Kirchhoff's voltage law,the source voltage $V_s$ is the su

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$2.5$

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(C) The diode and the resistor are connected in series with the voltage source.According to Kirchhoff's voltage law,the source voltage $V_s$ is the sum of the potential difference across the diode $(V_d)$ and the potential difference across the resistor $(V_r)$.Given:Potential difference across the diode,$V_d = 0.5\,V$Resistance,$R = 20\,\Omega$Current in the circuit,$I = 0.1\,A$The potential difference across the resistor is $V_r = I 	imes R = 0.1\,A 	imes 20\,\Omega = 2\,V$.Therefore,the source voltage is $V_s = V_d + V_r = 0.5\,V + 2\,V = 2.5\,V$.

$A$ diode having a potential difference of $0.5\,V$ across its junction,which does not depend on the current,is connected in series with a resistance of $20\,\Omega$ across a source. If $0.1\,A$ current passes through the resistance,what is the voltage of the source (in $,V$)?

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