In the given circuit, find the voltage drop V | vedclass

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(B) To determine if the Zener diode is conducting, we first calculate the voltage across the load resistance $R_L$ assuming the diode is open-circuite

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$3$

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(B) To determine if the Zener diode is conducting, we first calculate the voltage across the load resistance $R_L$ assuming the diode is open-circuited.Using the voltage divider rule, the voltage $V_{ab}$ across the load resistance $R_L = 100 \, \Omega$ is given by:$V_{ab} = V_{source} 	imes \frac{R_L}{R + R_L}$$V_{ab} = 9 \, V 	imes \frac{100 \, \Omega}{200 \, \Omega + 100 \, \Omega} = 9 \, V 	imes \frac{100}{300} = 3 \, V$Since the calculated voltage $V_{ab} = 3 \, V$ is less than the Zener breakdown voltage $V_Z = 5 \, V$, the Zener diode does not enter the breakdown region and remains in the non-conducting $(OFF)$ state.Therefore, the circuit behaves as a simple series circuit, and the voltage drop across the load resistance $R_L$ is $V_L = 3 \, V$.

In the given circuit, find the voltage drop $V_L$ across the load resistance $R_L$. (in $V$)

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