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(A) The circuit consists of a battery of $8 \, V$,a diode with a forward voltage drop of $0.5 \, V$,and a resistor of $2.2 \, k\Omega$ connected in se

Vedclass · Accepted Answer

$3.4$

Vedclass · Answer

(A) The circuit consists of a battery of $8 \, V$,a diode with a forward voltage drop of $0.5 \, V$,and a resistor of $2.2 \, k\Omega$ connected in series.Applying Kirchhoff's Voltage Law $(KVL)$ to the loop:$V_{battery} - V_{diode} - V_{resistor} = 0$$8 \, V - 0.5 \, V - I 	imes R = 0$$7.5 \, V = I 	imes (2.2 	imes 10^3 \, \Omega)$$I = \frac{7.5}{2.2 	imes 10^3} \, A$$I \approx 3.409 	imes 10^{-3} \, A$$I \approx 3.4 \, mA$Therefore,the current in the circuit is $3.4 \, mA$.

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

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