Question

When plate voltage in diode value is increased from $100V$to $150V$. Then, plate increases from $7.5\,mA$to $12\,mA$, then dynamic plate resistance will be
A. $10K\Omega$
B. $11K\Omega$
C. $15K\Omega$
D. $11.1K\Omega$

Answer 1

Hint In the question, the plate voltage in diode and plate current in diode is given. By using the expression of the dynamic plate resistance, we get the value of the dynamic plate resistance.
Formula used
The expression for finding the dynamic plate resistance is
${r_d} = \dfrac{{\Delta V}}{{\Delta I}}$
Where,
${r_d}$ be the dynamic resistance of a diode value, $\Delta V$ be the difference in change of the voltage, $\Delta I$ be the difference in change of the plate current.

Complete step by step solution
Given that,
Voltage ${V_1} = 100V$ and Voltage ${V_2} = 150V$
Current ${I_1} = 7.5mA$ and Current ${I_2} = 12mA$
Change in voltage ${V_2} - {V_1} = 150 - 100$
${V_2} - {V_1} = 50V$
Therefore, the change in voltage is $50V.$
Change in current ${I_2} - {I_1} = 12 - 7.5$
${I_2} - {I_1} = 4.5mA.$
${I_2} - {I_1} = 4.5 \times {10^{ - 3}}A.$
Therefore, the change in current is $4.5 \times {10^{ - 3}}A.$
The dynamic resistance of a diode value is the ratio of the small change in plate voltage to the small change in plate current. So, we written the expression as,
Dynamic plate resistance is ${r_d} = \dfrac{{\Delta V}}{{\Delta I}}$.
Substitute the known values in the above equation, we get
${r_d} = \dfrac{{50}}{{4.5 \times {{10}^{ - 3}}}}$
Performing the arithmetic operation in the above equation, we get
${r_d} = 11.1 \times {10^3}\Omega .$
Therefore, the value of the dynamic plate resistance is $11.1\,K\Omega .$

Hence, from the above options, option D is correct.

Note In the question, phase voltage of diode value is increased. But in the equation, we need the change in voltage and the current. So, we find the changing value of the current and the voltage. By substituting these values in the expression of dynamic plate resistance, we get the value of the diode resistance.