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 total source voltage $V$ is th

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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 total source voltage $V$ 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,$I = 0.1\, A$The potential difference across the resistor is $V_r = I 	imes R = 0.1\, A 	imes 20\, \Omega = 2\, V$.Therefore,the total source voltage is $V = V_d + V_r = 0.5\, V + 2\, V = 2.5\, V$.

$A$ 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 $0.1\, A$ passes through the resistance,then what is the voltage of the source (in $, V$)?

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