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Question

(A) Given: Voltage of the battery $V = 8\, V$,forward voltage drop across the diode $V_d = 0.5\, V$,and resistance $R = 2.2\, k\Omega = 2200\, \Omega$

Vedclass · Accepted Answer

$3.4$

Vedclass · Answer

(A) Given: Voltage of the battery $V = 8\, V$,forward voltage drop across the diode $V_d = 0.5\, V$,and resistance $R = 2.2\, k\Omega = 2200\, \Omega$.Applying Kirchhoff's Voltage Law $(KVL)$ in the loop:$V - V_d - I 	imes R = 0$$8 - 0.5 - I 	imes 2200 = 0$$7.5 = I 	imes 2200$$I = \frac{7.5}{2200}\, A$$I = 0.003409\, A$Converting to milliamperes $(mA)$:$I = 0.003409 	imes 1000\, mA = 3.409\, mA \approx 3.4\, mA$.Therefore,the correct option is $A$.

In the circuit,if the forward voltage drop for the diode is $0.5\, V$,the current will be.........$mA$.

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