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(B) In a Zener diode circuit,when the Zener diode is in the breakdown region,it maintains a constant voltage across itself equal to its Zener breakdow

Vedclass · Accepted Answer

$8$

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(B) In a Zener diode circuit,when the Zener diode is in the breakdown region,it maintains a constant voltage across itself equal to its Zener breakdown voltage $(V_Z)$.Given that the total voltage across the terminals $A$ and $B$ is $V_{total} = 17\,V$.The Zener breakdown voltage is $V_Z = 9\,V$.Since the resistor $R$ is in series with the Zener diode,the total voltage is the sum of the voltage across the resistor $(V_R)$ and the voltage across the Zener diode $(V_Z)$.Therefore,$V_{total} = V_R + V_Z$.Substituting the values: $17\,V = V_R + 9\,V$.Solving for $V_R$: $V_R = 17\,V - 9\,V = 8\,V$.Thus,the potential across $R$ is $8\,V$.

If the voltage between the terminals $A$ and $B$ is $17\,V$ and the Zener breakdown voltage is $9\,V$,then the potential across $R$ is....$V$

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